小编ftx*_*txn的帖子

http://在服务器中禁用了包装器?

错误消息:

Warning: include(): http:// wrapper is disabled in the server configuration by allow_url_include=0 in C:\xampp\htdocs\ubergallery\multiple_image_upload\multiupload.php on line 27

Warning: include(http://localhost/ubergallery/multiple_image_upload/upload.php): failed to open stream: no suitable wrapper could be found in C:\xampp\htdocs\ubergallery\multiple_image_upload\multiupload.php on line 27

Warning: include(): Failed opening 'http://localhost/ubergallery/multiple_image_upload/upload.php' for inclusion (include_path='.;C:\xampp\php\PEAR') in C:\xampp\htdocs\ubergallery\multiple_image_upload\multiupload.php on line 27
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我收到此错误消息.

包含带有绝对路径的样式表或jquery文件确实有效,我希望包含一个带有绝对路径的PHP脚本.

这是代码:

<!DOCTYPE html>
<html>
    <head>
        <title>Upload Multiple Images Using jquery and PHP</title>
        <!-------Including jQuery from google------>
        <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
        <script src="script.js"></script>

        <!-------Including CSS File------>
        <link rel="stylesheet" type="text/css" href="style.css">
    <body>
        <div id="maindiv"> …
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php apache

2
推荐指数
1
解决办法
2万
查看次数

jQuery - 路由是正确的,但警报不会发生?

index.html文件:

<!DOCTYPE html>
    <html>
<head>
    <title></title>
    <script src="jquery-2.1.1.min.js"></script>
    <script src="script.js"></script>
</head>
<body>
<h1>The Headline Tags Here</h1>
</body>
    </html>
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script.js文件:

$(document).ready(function() {

    alert("Hello!");


)};
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目录称为"灯箱":

index.html
jquery-2.1.1.min.js
script.js
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网址:

http://localhost/lightbox/
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我已经尝试通过你上面看到的方式加载jquery文件,然后我尝试用"警报"来检查它是否有效.虽然在页面加载时不会发生警报.

这是什么问题?

jquery

1
推荐指数
1
解决办法
44
查看次数

警告:explode()期望参数2是字符串,给定数组

剧本:

<?php

    $tqs = "SELECT * FROM `table_two`";
    $tqr = mysqli_query($dbc, $tqs);
    $row = mysqli_fetch_assoc($tqr);
    $thearray[] = $row['some_text_id'];

    // Prints e.g.: Array ( [0] => 164, 165, 166 )
    print_r($thearray);

    echo "<br/><br/>";
    echo "<br/><br/>";

    $thearray = explode(", ", $thearray);
    print_r($thearray);

?>
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我在"some_text_id"列的一行中有以下条目:

164, 165, 166
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我希望用逗号"爆炸"这个并将它存储在一个数组中,所以我可以单独选择数字,例如:

myarray[0], myarray[1], myarray[2]
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虽然我收到以下错误消息:

警告:explode()期望参数2是字符串,在...中给出的数组(指向爆炸函数)

有关如何做到这一点的任何建议?

php arrays explode

1
推荐指数
1
解决办法
3万
查看次数

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php ×2

apache ×1

arrays ×1

explode ×1

jquery ×1