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
– Bushra Shahid, Commented Apr 23, 2015 at 18:40

Community · Accepted Answer · 2017-05-23 11:51:02Z

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, 2017 at 11:51

CommunityBot

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answered Apr 23, 2015 at 18:46

Ninjakannon

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2 Comments

Travis Weston Over a year ago

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 Over a year ago

And with that edit, +1 for pretty much touching on every sane method of answering.

Alex · Accepted Answer · 2015-04-23 18:43:54Z

0

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

answered Apr 23, 2015 at 18:43

Alex

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Comments

Harmeet Singh Bhamra · Accepted Answer · 2015-04-23 18:46:31Z

0

Use concatenation like below....

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

Here, text in double quotes are string

answered Apr 23, 2015 at 18:46

Harmeet Singh Bhamra

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