How to add PHP variable in SELECT Query?

Question

I have a problem in add value in SELECT query.

$sql=("SELECT `image` FROM `testtable`");

The output: 123.jpg

But I want output: 127.0.0.1/home/galery/123.jpg

So I tried:

$path='127.0.0.1/home/galery/';
.........

$sql=("SELECT $path+`image` FROM `testtable`");

But it's not working.

if not then you should concatenate your output value with path .. — Bushra Shahid, Apr 23 '15 at 18:40

score 1 · Accepted Answer · edited May 23 '17 at 11:51

1

There are two ways to accomplish this.

Method 1:

Use string concatenation to join the path to the result from the SQL:

$path = '127.0.0.1/home/galery/';
$sql = "SELECT `image` FROM `testtable`";

// Run the query...

$result = $path . $sql;

In php, string concatenation is performed with the . operator. Also see here.

Method 2:

The second method is via the CONCAT SQL function:

$sql = "SELECT CONCAT('" . $path . "', `image`) FROM `testtable`";

Or:

$sql = "SELECT CONCAT('{$path}', `image`) FROM `testtable`";

See this question for the difference between these options.

edited May 23 '17 at 11:51

Community

answered Apr 23 '15 at 18:46

Ninjakannon

Your first example in Method 2 is incorrect, as wrapping `$path` in single quotes will make it parse as the literal `$path`, so you'd have to remove the single quotes or use the second example, which doesn't actually need the curly braces around $path, but those are useful for readability. – Travis Weston Apr 23 '15 at 19:08
And with that edit, +1 for pretty much touching on every sane method of answering. – Travis Weston Apr 23 '15 at 19:10

score 0 · Answer 2 · answered Apr 23 '15 at 18:43

0

$sql=("SELECT CONCAT('$path',`image`) FROM `testtable`");

answered Apr 23 '15 at 18:43

Alex

score 0 · Answer 3 · answered Apr 23 '15 at 18:46

0

Use concatenation like below....

$sql=("SELECT".$path."+image FROM test")

Here, text in double quotes are string

answered Apr 23 '15 at 18:46

3 Answers3