Home > Backend Development > PHP Tutorial > php中怎么输出javascript数组?

php中怎么输出javascript数组?

WBOY
Release: 2016-06-06 20:49:14
Original
1356 people have browsed it

php数组怎么输出成javascript中下面images值得样子?php数组中是url图片路径。 PHP数组格式为:

<code>Array([0] => http://XXX/xgsyw/content/Uploads/Img/1111.jpg[1] => http://XXX/xgsyw/content/Uploads/Img/222.jpg)

$('#top').bgStretcher({
            images: ['images/sample-1.jpg', 'images/sample-2.jpg', 'images/sample-3.jpg', 'images/sample-4.jpg', 'images/sample-5.jpg', 'images/sample-6.jpg'],
            slideDirection: 'N',
            slideShowSpeed: 1000,
            transitionEffect: 'fade',
            sequenceMode: 'normal',
        });
</code>
Copy after login
Copy after login

这个问题已被关闭,原因:

回复内容:

php数组怎么输出成javascript中下面images值得样子?php数组中是url图片路径。 PHP数组格式为:

<code>Array([0] => http://XXX/xgsyw/content/Uploads/Img/1111.jpg[1] => http://XXX/xgsyw/content/Uploads/Img/222.jpg)

$('#top').bgStretcher({
            images: ['images/sample-1.jpg', 'images/sample-2.jpg', 'images/sample-3.jpg', 'images/sample-4.jpg', 'images/sample-5.jpg', 'images/sample-6.jpg'],
            slideDirection: 'N',
            slideShowSpeed: 1000,
            transitionEffect: 'fade',
            sequenceMode: 'normal',
        });
</code>
Copy after login
Copy after login

你要是想输出JSON,php里json_encode是把数组转换成JSON,json_decode是把JSON转换成数组。肯定是可以用的,你可以说一下你遇到的具体问题。

你如果只是想输出你提到的

images: ['images/sample-1.jpg', 'images/sample-2.jpg', 'images/sample-3.jpg', 'images/sample-4.jpg', 'images/sample-5.jpg', 'images/sample-6.jpg']
那就拼一下字符串就好了

你给的那一段为什么会报错呢,那不就是申明一个数组么。回复里不能贴图,我就加在这里了。

php中怎么输出javascript数组?

拼接的话看看这样可以么?

<code>$array = ('a', 'b', 'c');
$array = array_map(create_function('$a', 'return "\"$a\"";'), $array);
$arr_js = '['.implde(',', $array).']';
</code>
Copy after login

不过其实可以考虑输出成json前段获取调用就是了。

<code><?php echo json_encode($images); ?>
</code>
Copy after login

是要输出Json字符串吗? 编码用json_encode($array) 解码用json_decode()

郁闷了,还是不成。到底怎么办呢?php啊

Related labels:
php
source:php.cn
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template