echo json_encode with PHP erroring

Question

Any idea why this is not working:

$result = mysql_query("SELECT cust_name, description from `projects` where project_no = '".$project_no."'") or die(mysql_error());
            while ($row = mysql_fetch_array($result)) {
                echo json_encode(
                      array("message1" => '".$row['cust_name']."', 
                      "message2" => '".$row['description']."')
                 )
            }

This is the error: Parse error: syntax error, unexpected T_STRING, expecting ')' in /admin/customerfilter.php on line 14

Line 14 is the line starting with "array..."

Thanks :)

Format your code properly, and you'll see what's wrong... Always indent and use a syntax-highlighting IDE/editor and you won't have this problem... — ircmaxell
– ircmaxell, Commented Jan 4, 2011 at 14:16

Mchl · Accepted Answer · 2011-01-04 14:06:27Z

4

Too many quotes:

$result = mysql_query("SELECT cust_name, description from `projects` where project_no = '".$project_no."'") or die(mysql_error());
            while ($row = mysql_fetch_array($result)) {
                echo json_encode(
                      array("message1" => $row['cust_name'], 
                      "message2" => $row['description'])
                 )
            }

answered Jan 4, 2011 at 14:06

Mchl

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2 Comments

teemitzitrone Over a year ago

add the missing ; and it's perfect

Mchl Over a year ago

I work just the way compiler does. Only fix the first error reported :P

Jon · Accepted Answer · 2011-01-04 14:07:37Z

3

echo json_encode(array(
    "message1" => $row['cust_name'],
    "message2" => $row['description'],
));

Don't make things more complicated than necessary.

answered Jan 4, 2011 at 14:07

Jon

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1 Comment

benhowdle89 Over a year ago

Thanks! perfect, the last thing i was trying to do was complicate it :) just get it to work!

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