$conf, $runtime; function_exists('chdir') AND chdir(APP_PATH); $r = 'mysql' == $conf['cache']['type'] ? website_set('runtime', $runtime) : cache_set('runtime', $runtime); } function runtime_truncate() { global $conf; 'mysql' == $conf['cache']['type'] ? website_set('runtime', '') : cache_delete('runtime'); } register_shutdown_function('runtime_save'); ?>How do I fix this: syntax error, unexpected ':', expecting ')'|Programmer puzzle solving

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    你的位置： HOME>matter>How do I fix this: syntax error, unexpected ':', expecting ')'

    How do I fix this: syntax error, unexpected ':', expecting ')'

    matteradmin9PV0评论

    Here is the line that is effecting my site from loading...I am just getting a blank page for my site. Please help as my business site is obviously down. Thank you in advance to whomever can help me.

    $bg = (!$bgimage == '') ? 'style="background-image:url'("".$bgimage[url].'")' : '';

    Here is the line that is effecting my site from loading...I am just getting a blank page for my site. Please help as my business site is obviously down. Thank you in advance to whomever can help me.

    $bg = (!$bgimage == '') ? 'style="background-image:url'("".$bgimage[url].'")' : '';

    • errors
    Share Improve this question asked Feb 13, 2022 at 21:20 Steve BonillaSteve Bonilla 1
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    1 Answer 1

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    Your code:

    $bg = (!$bgimage == '') ? 'style="background-image:url'("".$bgimage[url].'")' : '';

    There's a confusion with the quote marks. You're closing your ' too early, but it's really hard to see in a construction like this. (Also, the array index needs to be wrapped in quotes, either single or double.)

    $bg = (!$bgimage == '') ? 'style="background-image:url("'.$bgimage['url'].'")' : '';

    might clear up the PHP error, but I'm not sure it'll still do what you want it to. There's still some confusion with quote marks in the CSS; also, is $bgimage a string or an array?

    I'd recommend not using the ternary operator, in this case, to make the code more readable.

    $bg = '';
if ( ! empty( $bgimage ) ) {
    $bg = 'style="background-image:url(\'' . $bgimage['url'] . '\')";
}

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