Javascript 在 <script> 标签中使用 PHP？
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Using PHP in <script> tag?
提问by Gregor Menih
I'm trying to create an object in javascript with PHP. This is my script:
我正在尝试使用 PHP 在 javascript 中创建一个对象。这是我的脚本：
<script>
$(document).ready(function(){
var genres = [
<?php
$last = count($userGenres) - 1;
$i = 0;
foreach($userGenres as $a){
$i++;
echo '{"value":"'.$a['genre'].'", "name":"'.$a['name'].'"}';
if($i < $last){
echo ',';
}
}
?>
];
});
</script>
When I check the generated source, it creates a valid object, but the whole script in that tag doesn't work now. How would fix this, without JSON?
当我检查生成的源时，它创建了一个有效的对象，但该标记中的整个脚本现在不起作用。如果没有 JSON，如何解决这个问题？
Thanks
谢谢
采纳答案by Lars C. Magnusson
As the comment says you should consider using (php function)json_encode instead, which will turn...
正如评论所说，您应该考虑使用 (php function)json_encode 代替，这将变成...
php:
$a = array(
array("gender"=> "male", "name"=> "Bob"),
array("gender"=> "female", "name"=> "Annie")
);
Into json:
[
{gender:"male", name:"Bob"},
{gender:"female", name:"Annie"}
]
Echoing json_encode($a) would output it just like that.
回显 json_encode($a) 会像那样输出它。
回答by Kemal Fadillah
You forgot to close the .ready()method and the anonymous function inside of it.
您忘记关闭.ready()方法和其中的匿名函数。
Try this:
尝试这个：
<script>
$(document).ready(function(){
var genres = [
<?php
$last = count($userGenres) - 1;
$i = 0;
foreach($userGenres as $a)
{
$i++;
echo '{"value":"'.$a['genre'].'", "name":"'.$a['name'].'"}';
if($i < $last)
{
echo ',';
}
}
?>
];
});
</script>
回答by Bazzz
Your function is not closed properly:
您的功能未正确关闭：
}
}
?>
];
}); // <--- add this
</script>
回答by Mihalis Bagos
This:
这个：
<script>
$(document).ready(function(){
var genres = [
<?php
$last = count($userGenres) - 1;
$i = 0;
foreach($userGenres as $a)
{
echo '{"value":"'.$a['genre'].'", "name":"'.$a['name'].'"}';
if($i < $last)
{
echo ',';
}
$i++;
}
?>
];
});
</script>
Changes:You have to close the anonymous function and the ready() method. Also, move the i++ to the end so that you cover all cases (you will be missing one). The error you get is because you have the i++ wrong.
更改：您必须关闭匿名函数和 ready() 方法。此外，将 i++ 移到最后，以便涵盖所有情况（您将丢失一个）。你得到的错误是因为你的 i++ 错误。
