PHP Parse error: syntax error

Question

I have a mysqli_query like this:

SELECT a.* FROM (SELECT `id` as `id`, `age` as `age` FROM `register` WHERE `age` !="") as a INNER JOIN (SELECT `one` as `f1` FROM `friends` WHERE `two`='".$my_id."' UNION SELECT `two` as `f2` FROM `friends` WHERE `one` = '".$my_id."') as b ON a.id=b.f1

When I run this in my SQL it gives no error and shows a successful query but if I run it in my browser I get this error

( ! ) Parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING in C:\wamp\www\functions.php on line 585

Please anyone with an idea on how I can fix this?

This is my entire code

<?php
    $con = mysqli_connect('localhost','root','');

    mysqli_select_db($con, 'qings');

    $my_id = '1';
    // the line below is my line 585
    $query = mysqli_query($con, "SELECT a.* FROM (SELECT `id` as `id`, `age` as `age` FROM `register` WHERE `age` !="") as a INNER JOIN (SELECT `one` as `f1` FROM `friends` WHERE `two`='".$my_id."' UNION SELECT `two` as `f2` FROM `friends` WHERE `one` = '".$my_id."') as b ON a.id=b.f1");

?>

Show the relevant PHP code, the error is not in the SQL query yet — LeleDumbo, Jul 03 '16 at 11:07
You need to post the PHP code that generates the error. It's in `C:\wamp\www\functions.php` on line 585, as the error message shows. — mhawke, Jul 03 '16 at 11:08
The line 585 is the query itself which is in the question. I have shown my full code and you can see it yourself — KANAYO AUGUSTIN UG, Jul 03 '16 at 11:27
@Chay22 ... Thanks so much, I removed it and the error is gone :-) — KANAYO AUGUSTIN UG, Jul 03 '16 at 11:32

score 1 · Accepted Answer · answered Jul 03 '16 at 12:25

Change WHERE 'age' !="" to WHERE 'age' !=''

Use

$query = mysqli_query($con, "SELECT a.* FROM (SELECT `id` as `id`, `age` as `age` FROM `register` WHERE `age` !='') as a INNER JOIN (SELECT `one` as `f1` FROM `friends` WHERE `two`='".$my_id."' UNION SELECT `two` as `f2` FROM `friends` WHERE `one` = '".$my_id."') as b ON a.id=b.f1");

score 0 · Answer 2 · edited Jul 03 '16 at 12:28

0

You should just escape double quotes here

WHERE `age` != \"\"

edited Jul 03 '16 at 12:28

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answered Jul 03 '16 at 12:05

Nick

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PHP Parse error: syntax error

2 Answers2