仅当结果中存在数组时如何回显 str_replace

Question

我在 PHP 方面非常业余，所以希望这是有道理的。我有以下 foreach 循环，它使用艺术家的 id（混音器、制作人、编曲者、作曲家等）获取链接到音乐艺术家的所有 "credits"。我希望能够用更多的东西替换不同的信用名称 "list friendly." 下面的尝试有效，但它只会循环，直到它遇到与艺术家无关的信用，然后它停止。

<?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; ?>

如何将 if 语句添加到代码中，以便它仅替换结果中的 "array"？

Answer 1

添加检查 $skill['credit_name'] 是否为数组 -

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

Answer 2

您可以使用 is_array function along with 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']);
}

Answer 3

除了在 foreach() 中评估技能外，你没有做任何事情，你可以将整个块放在外面，如下所示，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?>

Answer 4

感谢大家的回复，我摆弄了张贴的建议，意识到我没有在第一个数组中指定足够多的原始 credit_name 来匹配第二个数组。而且它也必须是一个字符串，因为 if(is_string) 有效而 if(is_array) 无效。（正如我所说，所有这些仍然非常环保！）

<?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; ?>

Answer 5

你可以这样做，

$result = getSkills($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。

仅当结果中存在数组时如何回显 str_replace

How to echo an str_replace only if array exists in results

php

foreach

echo

str-replace