Use PHP variable in a SELECT LIKE query
I'm trying to execute a mysqli query using the following code:
$sql = "SELECT * FROM `table` WHERE `description` LIKE ('AB CD %')";
$result= mysqli_query($con, $sql) or die(mysqli_error());
and this query gives me 6 results. It will only looks for items like "AB CD EF..." and not items like "AB CDEF...", which is exactly what I want.
But if I give to the LIKE value a variable like this:
$var = "AB CD ";
$sql = "SELECT * FROM `table` WHERE `description` LIKE ('$var%')";
$result= mysqli_query($con, $sql) or die(mysqli_error());
it gives me zero results.
I have tried also several LIKE formats such ...('".$var."%')"; or CONCAT($var, '%'), but nothing.
How can get the same results as the first query usin a variable like in the sencond query?
The variable is get by a query which will select all description items and then, inside a while loop, it will look for the first Capital letters of each item:
$name = $row['description'];
$expr = '/[A-Z]*/';
preg_match_all($expr, $name, $res);
$var = implode(' ', $res[0]);
Each row has values like "AB CD EF something else in not capital letters"
Thank you.
php mysqli
asked Nov 24 '18 at 16:26
GoncalorsrGoncalorsr
134
|
show 12 more comments
I'm trying to execute a mysqli query using the following code:
$sql = "SELECT * FROM `table` WHERE `description` LIKE ('AB CD %')";
$result= mysqli_query($con, $sql) or die(mysqli_error());
and this query gives me 6 results. It will only looks for items like "AB CD EF..." and not items like "AB CDEF...", which is exactly what I want.
But if I give to the LIKE value a variable like this:
$var = "AB CD ";
$sql = "SELECT * FROM `table` WHERE `description` LIKE ('$var%')";
$result= mysqli_query($con, $sql) or die(mysqli_error());
it gives me zero results.
I have tried also several LIKE formats such ...('".$var."%')"; or CONCAT($var, '%'), but nothing.
How can get the same results as the first query usin a variable like in the sencond query?
The variable is get by a query which will select all description items and then, inside a while loop, it will look for the first Capital letters of each item:
$name = $row['description'];
$expr = '/[A-Z]*/';
preg_match_all($expr, $name, $res);
$var = implode(' ', $res[0]);
Each row has values like "AB CD EF something else in not capital letters"
Thank you.
php mysqli
asked Nov 24 '18 at 16:26
GoncalorsrGoncalorsr
134
how about$var = 'AB CD %';?
– Always Sunny
Nov 24 '18 at 16:28
Yes, tried, also.
– Goncalorsr
Nov 24 '18 at 16:31
Maybe is important to refer how I get the variable. I will edit the post
– Goncalorsr
Nov 24 '18 at 16:31
Yes that is very much important. I guess you need to escape the string also.
– Always Sunny
Nov 24 '18 at 16:32
2
How about using prepared statements...
– dn Fer
Nov 24 '18 at 16:35
|
show 12 more comments
I'm trying to execute a mysqli query using the following code:
$sql = "SELECT * FROM `table` WHERE `description` LIKE ('AB CD %')";
$result= mysqli_query($con, $sql) or die(mysqli_error());
and this query gives me 6 results. It will only looks for items like "AB CD EF..." and not items like "AB CDEF...", which is exactly what I want.
But if I give to the LIKE value a variable like this:
$var = "AB CD ";
$sql = "SELECT * FROM `table` WHERE `description` LIKE ('$var%')";
$result= mysqli_query($con, $sql) or die(mysqli_error());
it gives me zero results.
I have tried also several LIKE formats such ...('".$var."%')"; or CONCAT($var, '%'), but nothing.
How can get the same results as the first query usin a variable like in the sencond query?
The variable is get by a query which will select all description items and then, inside a while loop, it will look for the first Capital letters of each item:
$name = $row['description'];
$expr = '/[A-Z]*/';
preg_match_all($expr, $name, $res);
$var = implode(' ', $res[0]);
Each row has values like "AB CD EF something else in not capital letters"
Thank you.
php mysqli
asked Nov 24 '18 at 16:26
GoncalorsrGoncalorsr
134
I'm trying to execute a mysqli query using the following code:
$sql = "SELECT * FROM `table` WHERE `description` LIKE ('AB CD %')";
$result= mysqli_query($con, $sql) or die(mysqli_error());
and this query gives me 6 results. It will only looks for items like "AB CD EF..." and not items like "AB CDEF...", which is exactly what I want.
But if I give to the LIKE value a variable like this:
$var = "AB CD ";
$sql = "SELECT * FROM `table` WHERE `description` LIKE ('$var%')";
$result= mysqli_query($con, $sql) or die(mysqli_error());
it gives me zero results.
I have tried also several LIKE formats such ...('".$var."%')"; or CONCAT($var, '%'), but nothing.
How can get the same results as the first query usin a variable like in the sencond query?
The variable is get by a query which will select all description items and then, inside a while loop, it will look for the first Capital letters of each item:
$name = $row['description'];
$expr = '/[A-Z]*/';
preg_match_all($expr, $name, $res);
$var = implode(' ', $res[0]);
Each row has values like "AB CD EF something else in not capital letters"
Thank you.
php mysqli
php mysqli
asked Nov 24 '18 at 16:26
GoncalorsrGoncalorsr
134
asked Nov 24 '18 at 16:26
GoncalorsrGoncalorsr
134
edited Nov 24 '18 at 16:39
Goncalorsr
asked Nov 24 '18 at 16:26
GoncalorsrGoncalorsr
134
asked Nov 24 '18 at 16:26
GoncalorsrGoncalorsr
134
asked Nov 24 '18 at 16:26
GoncalorsrGoncalorsr
134
134
how about$var = 'AB CD %';?
– Always Sunny
Nov 24 '18 at 16:28
Yes, tried, also.
– Goncalorsr
Nov 24 '18 at 16:31
Maybe is important to refer how I get the variable. I will edit the post
– Goncalorsr
Nov 24 '18 at 16:31
Yes that is very much important. I guess you need to escape the string also.
– Always Sunny
Nov 24 '18 at 16:32
2
How about using prepared statements...
– dn Fer
Nov 24 '18 at 16:35
|
show 12 more comments
how about$var = 'AB CD %';?
– Always Sunny
Nov 24 '18 at 16:28
Yes, tried, also.
– Goncalorsr
Nov 24 '18 at 16:31
Maybe is important to refer how I get the variable. I will edit the post
– Goncalorsr
Nov 24 '18 at 16:31
Yes that is very much important. I guess you need to escape the string also.
– Always Sunny
Nov 24 '18 at 16:32
2
How about using prepared statements...
– dn Fer
Nov 24 '18 at 16:35
how about
$var = 'AB CD %'; ?– Always Sunny
Nov 24 '18 at 16:28
how about
$var = 'AB CD %'; ?– Always Sunny
Nov 24 '18 at 16:28
Yes, tried, also.
– Goncalorsr
Nov 24 '18 at 16:31
Yes, tried, also.
– Goncalorsr
Nov 24 '18 at 16:31
Maybe is important to refer how I get the variable. I will edit the post
– Goncalorsr
Nov 24 '18 at 16:31
Maybe is important to refer how I get the variable. I will edit the post
– Goncalorsr
Nov 24 '18 at 16:31
Yes that is very much important. I guess you need to escape the string also.
– Always Sunny
Nov 24 '18 at 16:32
Yes that is very much important. I guess you need to escape the string also.
– Always Sunny
Nov 24 '18 at 16:32
2
2
How about using prepared statements...
– dn Fer
Nov 24 '18 at 16:35
How about using prepared statements...
– dn Fer
Nov 24 '18 at 16:35
|
show 12 more comments
3 Answers
3
active
oldest
votes
How about this after wrapping the variable with curly braces {}? Also I just removed the extra parenthesis () after LIKE :)
$var = "AB CD ";
$sql = "SELECT * FROM `table` WHERE `description` LIKE '{$var}%'";
answered Nov 24 '18 at 16:31
Always SunnyAlways Sunny
16.4k32847
Yes I tried this, also.
– Goncalorsr
Nov 24 '18 at 16:39
add a comment |
Well, user3783243 found the problem.
When I implode the array to get the $var in a string, for some reason I cannot explain, the array elements (a lot of them) that had no value were included. The solution was to explode the $var, pass it through the array_filter function, to delete the empty elements and implode it again.
Thank you all.
answered Nov 24 '18 at 18:22
GoncalorsrGoncalorsr
134
add a comment |
Use :
$var = "AB CD ";
$sql = "SELECT * FROM `table` WHERE `description` LIKE '".$var."%'";
But this is a bad practice as well as highly vulnerable, use prepared statement instead : http://php.net/manual/en/book.pdo.php
answered Nov 24 '18 at 17:57
LabLab
1,00489
That is the same as what the OP has. PDO can be just as secure/insecure as mysqli.
– user3783243
Nov 24 '18 at 18:06
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
How about this after wrapping the variable with curly braces {}? Also I just removed the extra parenthesis () after LIKE :)
$var = "AB CD ";
$sql = "SELECT * FROM `table` WHERE `description` LIKE '{$var}%'";
answered Nov 24 '18 at 16:31
Always SunnyAlways Sunny
16.4k32847
Yes I tried this, also.
– Goncalorsr
Nov 24 '18 at 16:39
add a comment |
How about this after wrapping the variable with curly braces {}? Also I just removed the extra parenthesis () after LIKE :)
$var = "AB CD ";
$sql = "SELECT * FROM `table` WHERE `description` LIKE '{$var}%'";
answered Nov 24 '18 at 16:31
Always SunnyAlways Sunny
16.4k32847
Yes I tried this, also.
– Goncalorsr
Nov 24 '18 at 16:39
add a comment |
How about this after wrapping the variable with curly braces {}? Also I just removed the extra parenthesis () after LIKE :)
$var = "AB CD ";
$sql = "SELECT * FROM `table` WHERE `description` LIKE '{$var}%'";
answered Nov 24 '18 at 16:31
Always SunnyAlways Sunny
16.4k32847
How about this after wrapping the variable with curly braces {}? Also I just removed the extra parenthesis () after LIKE :)
$var = "AB CD ";
$sql = "SELECT * FROM `table` WHERE `description` LIKE '{$var}%'";
answered Nov 24 '18 at 16:31
Always SunnyAlways Sunny
16.4k32847
answered Nov 24 '18 at 16:31
Always SunnyAlways Sunny
16.4k32847
answered Nov 24 '18 at 16:31
Always SunnyAlways Sunny
16.4k32847
answered Nov 24 '18 at 16:31
Always SunnyAlways Sunny
16.4k32847
16.4k32847
Yes I tried this, also.
– Goncalorsr
Nov 24 '18 at 16:39
add a comment |
Yes I tried this, also.
– Goncalorsr
Nov 24 '18 at 16:39
Yes I tried this, also.
– Goncalorsr
Nov 24 '18 at 16:39
Yes I tried this, also.
– Goncalorsr
Nov 24 '18 at 16:39
add a comment |
Well, user3783243 found the problem.
When I implode the array to get the $var in a string, for some reason I cannot explain, the array elements (a lot of them) that had no value were included. The solution was to explode the $var, pass it through the array_filter function, to delete the empty elements and implode it again.
Thank you all.
answered Nov 24 '18 at 18:22
GoncalorsrGoncalorsr
134
add a comment |
Well, user3783243 found the problem.
When I implode the array to get the $var in a string, for some reason I cannot explain, the array elements (a lot of them) that had no value were included. The solution was to explode the $var, pass it through the array_filter function, to delete the empty elements and implode it again.
Thank you all.
answered Nov 24 '18 at 18:22
GoncalorsrGoncalorsr
134
add a comment |
Well, user3783243 found the problem.
When I implode the array to get the $var in a string, for some reason I cannot explain, the array elements (a lot of them) that had no value were included. The solution was to explode the $var, pass it through the array_filter function, to delete the empty elements and implode it again.
Thank you all.
answered Nov 24 '18 at 18:22
GoncalorsrGoncalorsr
134
Well, user3783243 found the problem.
When I implode the array to get the $var in a string, for some reason I cannot explain, the array elements (a lot of them) that had no value were included. The solution was to explode the $var, pass it through the array_filter function, to delete the empty elements and implode it again.
Thank you all.
answered Nov 24 '18 at 18:22
GoncalorsrGoncalorsr
134
answered Nov 24 '18 at 18:22
GoncalorsrGoncalorsr
134
answered Nov 24 '18 at 18:22
GoncalorsrGoncalorsr
134
answered Nov 24 '18 at 18:22
GoncalorsrGoncalorsr
134
134
add a comment |
add a comment |
Use :
$var = "AB CD ";
$sql = "SELECT * FROM `table` WHERE `description` LIKE '".$var."%'";
But this is a bad practice as well as highly vulnerable, use prepared statement instead : http://php.net/manual/en/book.pdo.php
answered Nov 24 '18 at 17:57
LabLab
1,00489
That is the same as what the OP has. PDO can be just as secure/insecure as mysqli.
– user3783243
Nov 24 '18 at 18:06
add a comment |
Use :
$var = "AB CD ";
$sql = "SELECT * FROM `table` WHERE `description` LIKE '".$var."%'";
But this is a bad practice as well as highly vulnerable, use prepared statement instead : http://php.net/manual/en/book.pdo.php
answered Nov 24 '18 at 17:57
LabLab
1,00489
That is the same as what the OP has. PDO can be just as secure/insecure as mysqli.
– user3783243
Nov 24 '18 at 18:06
add a comment |
Use :
$var = "AB CD ";
$sql = "SELECT * FROM `table` WHERE `description` LIKE '".$var."%'";
But this is a bad practice as well as highly vulnerable, use prepared statement instead : http://php.net/manual/en/book.pdo.php
answered Nov 24 '18 at 17:57
LabLab
1,00489
Use :
$var = "AB CD ";
$sql = "SELECT * FROM `table` WHERE `description` LIKE '".$var."%'";
But this is a bad practice as well as highly vulnerable, use prepared statement instead : http://php.net/manual/en/book.pdo.php
answered Nov 24 '18 at 17:57
LabLab
1,00489
answered Nov 24 '18 at 17:57
LabLab
1,00489
answered Nov 24 '18 at 17:57
LabLab
1,00489
answered Nov 24 '18 at 17:57
LabLab
1,00489
1,00489
That is the same as what the OP has. PDO can be just as secure/insecure as mysqli.
– user3783243
Nov 24 '18 at 18:06
add a comment |
That is the same as what the OP has. PDO can be just as secure/insecure as mysqli.
– user3783243
Nov 24 '18 at 18:06
That is the same as what the OP has. PDO can be just as secure/insecure as mysqli.
– user3783243
Nov 24 '18 at 18:06
That is the same as what the OP has. PDO can be just as secure/insecure as mysqli.
– user3783243
Nov 24 '18 at 18:06
add a comment |
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how about
$var = 'AB CD %';?– Always Sunny
Nov 24 '18 at 16:28
Yes, tried, also.
– Goncalorsr
Nov 24 '18 at 16:31
Maybe is important to refer how I get the variable. I will edit the post
– Goncalorsr
Nov 24 '18 at 16:31
Yes that is very much important. I guess you need to escape the string also.
– Always Sunny
Nov 24 '18 at 16:32
2
How about using prepared statements...
– dn Fer
Nov 24 '18 at 16:35