如何只在array存在于结果中时回显str_replace - How to echo an str_replace only if array exists in results

对于PHP我是非常业余的，所以希望这是有意义的。我有以下foreach循环，获取所有的"学分"，链接到一个音乐艺术家使用该艺术家的id(混音，制作人，编曲，作曲家等)。我希望能够用更"列表友好"的东西替换不同的信用名称。下面的尝试是有效的，但它只能循环，直到它碰到一个与艺术家无关的字幕，然后它就停止了。

<?php function getSkills($id)
{
    $query = "SELECT c2a.credit_id, cr.credit_name
FROM  `Credit_To_Artist` AS c2a
INNER JOIN  `Credits` AS cr ON cr.credit_id = c2a.credit_id
INNER JOIN  `Artist` AS a ON a.artist_id = c2a.artist_id
WHERE c2a.artist_id = $id
GROUP BY c2a.credit_id
ORDER BY cr.credit_name";
    $res = mysql_query($query);
    while ($row = mysql_fetch_assoc($res)) {
        $skills[] = $row;
    }
    return $skills;
}
?>
<?php foreach (getSkills($id) as $skill): ?>
    <?php echo str_replace(
            array('arranger','mixer','producer','composer','engineer','mixer','recorder','vocalist','writer'),
            array('song arrangement','audio mixing','music production','music/vocal recording','song writing','singing/performance'),
            $skill['credit_name']); ?><br />
<?php endforeach; ?>

如何在代码中添加if语句，使其仅替换结果中的"数组"?

添加检查$skill['credit_name']是否为数组-

if(is_array($skill['credit_name'])) {
    // Your code
}

您可以使用is_array函数和count作为

if(is_array($skill['credit_name']) && count($skill['credit_name']) > 0){
    echo str_replace(array('arranger','mixer','producer','composer','engineer','mixer','recorder','vocalist','writer'), array('song arrangement','audio mixing','music production','music/vocal recording','song writing','singing/performance'), $skill['credit_name']);
}

您除了在此foreach()中评估技能之外没有做任何事情，您可以将整个块放在下面，假设$skill['credit_name']是字符串，

<?php if(is_array(getSkills($id)): ?>
<?php foreach (getSkills($id) as $skill): ?>
<?php if(is_string($skill['credit_name']) && $skill['credit_name']!=''): ?>

<?php echo str_replace(
        array('arranger','mixer','producer','composer','engineer','mixer','recorder','vocalist','writer'),
        array('song arrangement','audio mixing','music production','music/vocal recording','song writing','singing/performance'),
        $skill['credit_name']); ?><br />
<?php endif?>
<?php endforeach?>
<?php endif?>

感谢每个人的回复，我处理了发布的建议，并意识到我没有在第一个数组中指定足够的原始credit_name 's来匹配第二个数组。它也必须是字符串，因为if(is_string)可以工作，if(is_array)不行。(就像我说的，这一切还是很绿色的!)

<?php foreach (getSkills($id) as $skill): ?>
    <?php if(is_string($skill['credit_name'])) {
    echo str_replace(
            array('arranger','mixer','producer','composer','engineer','recorder','vocalist','writer'),
            array('song arrangement','audio mixing','music production','song composition','music engineering','music/vocal recording','singing/performance','song writing'),
            $skill['credit_name']);
    echo "<br />";
    } ?>
<?php endforeach; ?>

您可以这样做，

$result = getSkills($id);
$search = array('arranger','mixer','producer','composer','engineer','mixer','recorder','vocalist','writer');
$replace = array('song arrangement','audio mixing','music production','music/vocal recording','song writing','singing/performance');
foreach($result as &$v)
{
    $v['credit_name'] = str_replace($search,$replace,$v['credit_name']);
}

现在$result包含替换的字符串