PHP variable in WHERE syntax issue :(

Question

So, I am trying to do a simple query which works fine when I put in the URL manually as a string. But when I put it as a variable it doesn't work. I even checked the variable and the variable has the URl in it. Never put a variable in a WHERE statement so not sure if it is an echo thing or what, just that nothing seems to work.

$external_result = $facebook->api(
    array(
        'method' => 'fql.query',
        'query' => 'SELECT share_count, 
                           like_count, 
                           comment_count, 
                           total_count 
                      FROM link_stat 
                     WHERE url="{$urlf}";'
    )
);

echo $external_result[0]['total_count'];

Sherif · Accepted Answer · 2012-12-10 20:37:48Z

1

There's a difference between single and double quoted strings in PHP syntax. A single quoted string does not offer variable interpolation. Whereas double quoted strings do allow you to interpolate variables inside string literals. See the manual on PHP strings for more details.

Here's what you need to do.

$external_result = $facebook->api(
    array(
        'method' => 'fql.query',
        'query' => "SELECT share_count, like_count, comment_count, total_count FROM link_stat WHERE url=\"{$urlf}\";"
    )
);

By replacing your single quotes around that string to double quotes you now get PHP to replace $urlf with it's actual value.

answered Dec 10, 2012 at 20:37

Sherif

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Comments

user554546 · Accepted Answer · 2012-12-10 20:35:29Z

1

Single-quoted strings do not allow for variable interpolation:

$foo = 5;
echo("This is $foo"); # "This is 5"
echo('This is ' . $foo); #ditto
echo('This is $foo'); # "This is $foo"
echo("This is \$foo"); # ditto

I'd recommend looking through some PHP tutorials or picking up a book.

answered Dec 10, 2012 at 20:35

user554546

Comments

anubhava · Accepted Answer · 2012-12-10 21:39:47Z

0

Change your code to:

$external_result = $facebook->api(
    array(
        'method' => 'fql.query',
        'query' => 'SELECT share_count, like_count, comment_count, 
                    total_count FROM link_stat WHERE url="' . $urlf . '";'
    )
);

edited Dec 10, 2012 at 21:39

answered Dec 10, 2012 at 20:35

anubhava

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3 Comments

Nick Over a year ago

I tried this but it came up with no value, but the double quotes issue just above fixed it. Hmmmm.

anubhava Over a year ago

Its your prerogative to mark any answer as accepted however just to let you know that there is no difference between your accepted answer and mine other than I am using much preferred way of concatenating variable with string in PHP.

anubhava Over a year ago

@Neal: Thanks, yes I see, removed curly braces.

Naftali · Accepted Answer · 2012-12-10 20:35:13Z

-1

Use double quotes if you are going to do something like that.

'query' => "
            SELECT share_count, like_count, comment_count, total_count 
            FROM link_stat WHERE url='{$urlf}';
           "

answered Dec 10, 2012 at 20:35

Naftali

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2 Comments

Nick Over a year ago

This totally worked but you said don't! Crap. $external_result = $facebook->api(array( 'method' => 'fql.query', 'query' => "SELECT share_count, like_count, comment_count, total_count FROM link_stat WHERE url='{$urlf}';" )); echo $external_result[0]['total_count'];

Madara's Ghost Over a year ago

Not sure why this got downvoted either. It's a perfectly acceptable answer, which indeed solves the problem.

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