我正在尝试在MySQL表中插入Cyrillic值,但编码存在问题.
PHP的:
<?php
$servername = "localhost";
$username = "a";
$password = "b";
$dbname = "c";
$conn = new mysqli($servername, $username, $password, $dbname);
mysql_query("SET NAMES 'utf8';");
mysql_query("SET CHARACTER SET 'utf8';");
mysql_query("SET SESSION collation_connection = 'utf8_general_ci';");
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "UPDATE `c`.`mainp` SET `search` = 'test ????' WHERE `mainp`.`id` =1;";
if ($conn->query($sql) === TRUE) {
}
$conn->close();
?>
Run Code Online (Sandbox Code Playgroud)
MySQL的:
| id | search |
| 1 | test ав |
Run Code Online (Sandbox Code Playgroud)
注意:PHP文件是utf-8,数据库排序规则utf8_general_ci
我有7个html div.如何使用data-recentviewes作为每个元素的load-URL,仅为前6个div加载一些动态数据?
HTML:
<div id="recentViewes" data-recentviewes="/123/"></div>
<div id="recentViewes" data-recentviewes="/456/"></div>
<div id="recentViewes" data-recentviewes="/789/"></div>
<div id="recentViewes" data-recentviewes="/321/"></div>
<div id="recentViewes" data-recentviewes="/654/"></div>
<div id="recentViewes" data-recentviewes="/987/"></div> <!-- 6 -->
---
<div id="recentViewes" data-recentviewes="/abc/"></div> <!-- dont load this -->
Run Code Online (Sandbox Code Playgroud)
js(仅适用于第一个元素):
window.addEventListener('load', function () {
var recentviewes = $("#recentViewes").data('recentviewes');
$("#recentViewes").each(function(index, element) {
$("#recentViewes").load("http://example.com" + recentviewes);
return index < 5;
});
});
Run Code Online (Sandbox Code Playgroud)
谢谢.
<div id="id1">
apple
ball
dogsss
dogsssdogsss
dogsssdogsssdogsss
</div>
Run Code Online (Sandbox Code Playgroud)
如何将 ALL 更改dogsss为dollsss使用 jquery?
这是一个代码,但如何确定元素?
$('#id1 how???').each(function() {
var text = $(this).text();
$(this).text(text.replace('dog', 'doll'));
});
Run Code Online (Sandbox Code Playgroud)
这是类似的问题 http://stackoverflow.com/questions/8146648/jquery-find-text-and-replace
我有一个带图像的文件夹.
例如:
z_1.jpg
z_2.jpg
z_3.jpg
//...
Run Code Online (Sandbox Code Playgroud)
我想用前缀删除每个图像z_*.jpg.我怎样才能做到这一点?
unlink('z_*.jpg'); ?
Run Code Online (Sandbox Code Playgroud)