PHP variable combining issue

Question

I am trying to concatenate a variable to my sql query in php to have my table sorted according to that variable. For eg:

if ($_GET['sort'] == 'Val'){
   $query .= " ORDER BY val";
}

I need a variabvle $sortorder to be concatenated with this query. so that i get sorted query in ASC or DESC. What i do is

$order = ASC ;

if ($_GET['sort'] == 'Val'){
   $query .= " ORDER BY val" ."$order";
}

and write the logic for

if($order==ASC){$order= DESC}

and vice versa. but it doesnt work. table is not displayed. some syntax issue?

Anyone could help on this. ?

What error do you get? What did you do to troubleshoot this? — John Conde
– John Conde, Commented Sep 19, 2013 at 1:21
"some syntax issue" - what is the issue? Any PHP or mysql errors? — James
– James, Commented Sep 19, 2013 at 1:21
If they are not constants, you should wrap them in quotes. $order = 'ASC'; — Dave Chen
– Dave Chen, Commented Sep 19, 2013 at 1:24

Mayank K. Swami · Accepted Answer · 2013-09-19 02:14:47Z

2

You need to introduce a space. Otherwise SQL will read ORDER BY valDESC

So:

$query .= " ORDER BY val " ."$order";

Or more compact:

$query .= " ORDER BY val $order";

You can check that your query is as expected by simply printing it

echo $query ;

This simple debugging can help you to resolve problem

edited Sep 19, 2013 at 2:14

Mayank K. Swami

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answered Sep 19, 2013 at 1:24

willeM_ Van Onsem

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Comments

Barmar · Accepted Answer · 2013-09-19 01:24:22Z

0

Use:

$query .= " ORDER BY val $order";

You need a space between the column name and the sort order. Your code was generating:

ORDER BY valASC

If you simply did echo $query; before trying to execute it, you would surely have seen this. Please do some basic, obvious debugging before posting here.

answered Sep 19, 2013 at 1:24

Barmar

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1 Comment

user2708092 Over a year ago

Yes i resolved this now. I also need, once it is sorted in ASC, next time on click it is DESc. any points on that.. Thank you for your response

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