你好,我已经面对这个问题已经好几天了,我找不到办法解决这个问题,或者围绕它解决这个问题。
我想要做的很简单,我想读出一个大项目文件夹的每个子文件夹。然后将缩略图图像和图像分配到此文件夹。通过一个简单的for循环,php为我构建了这个。一切都很完美和快速。唯一的问题是jquery不会响应。
即使我用这种技术创造了各种菜单。正如您在我的代码中看到的那样,在“脚本”标签中,我有jquery代码似乎不起作用。
我不知道php放在某个地方的某个空间,或者我只是看了太长时间看这个代码来查看错误。
我感谢任何帮助。
<?php
/*Because of the "ease of use" aspect of this site, I prefered to write it completely in PHP,
advantage of that:
Images are loaded directly out of the folder, so I can just drag and drop something onto the server without having to write aditional code.
As you see, it can save a lot of time.*/
echo "<h1>Referenzen</h1><br>";
$projects = scandir('../src/sub/credentials'); //The credentials directory is being scanned and converted into an array.
$projectsSize = count($projects); //$size is being created. It counts the number of objects in the array which has been created above.
$projectsCaptions = file('../src/sub/captionsOfCredentials.txt'); //Edit the name of the figcaption in the "captionsOfCredentials.txt".
for($i = 2; $i < $projectsSize; $i++){ /*Simple "for" loop, that creates $i with the size of two, because PHP is 0-index based and has the "dot" and the "dotdot" folder. The loop stops at the end of the array($size).*/
echo '<a href="index.php#PRJ'.trim($projectsCaptions[$i]).'" class="ID'.trim($projectsCaptions[$i]).'">
<div class="projectFolder">
<img src="src/sub/credentialsThumb/project_00'.$i.'.jpg" width="100%" />
<figcaption>'
.$projectsCaptions[$i].
'</figcaption>
</div>
</a>';
/*Project folder level starts here.*/
$images = scandir('../src/sub/credentials/project_00'.$i);
$imagesSize = count($images);
for($k = 3; $k < $imagesSize; $k++){
$tempID = ('ID'.trim($projectsCaptions[$i]).'.php'); //$tempID is the entire file name including the .php part.
$handle = fopen($tempID, "a") or die ('Unable to open '.$tempID.' , please contact admin A.S.A.P..');
$imagesCode = 'test';
fwrite($handle, $imagesCode);
}
//end second for-loop
echo "
<script>
$(document).ready(function () {
$('#ID".$projectsCaptions[$i]."').click(function () {
$('#mainContent').load('de-DE/".$tempID."');
});
});
</script>";
}
//end first for-loop
?>
答案 0 :(得分:1)
当您需要使用类时,您将通过 id 选择元素。将JS块更改为:
$(document).ready(function () {
// Note ".ID" not "#ID"
$('.ID".$projectsCaptions[$i]."').click(function () {
$('#mainContent').load('de-DE/".$tempID."');
});
});
<强>更新强>
好像你在$projectsCaptions[$i]
中也有一个非法角色。它很可能是换行符。尝试在trim()
:
$('.ID" . trim($projectsCaptions[$i]) . "').click(function () {