[英]MySQL, PHP Using LIKE Syntax?
I'm trying to make a user search with the following code: 我正在尝试使用以下代码进行用户搜索:
<?php
session_start();
include("../BD/bd.php");
$searched_for = $_POST['searched_for'];
$query = @mysql_query("SELECT * FROM user_media WHERE nombre LIKE '%$searched_for%'") or die(mysql_error());
while($got_users = @mysql_fetch_array($query)){
echo '<div class="searched-content-info">'.
'<div class="searched-photo"><img src="'.$got_users['foto'].'"></div>
<div class="searched-names"><h3>'.$got_users['nombre'].'</h3></div>
<div class="searched-dates"><h3>'.'Miembro desde: '.$got_users['created_on'].'</h3></div>
</div>
<div class="divisor-search-user"></div>';
}
?>
But I'm getting all the rows, I just want to display the searched users info, seems like the $query
is receiving a clean $searched_for
但是我正在获取所有行,我只想显示搜索到的用户信息,似乎
$query
正在接收干净的$searched_for
Any help here? 这里有什么帮助吗? Btw, I'm a little newbie here, please don't bully :)
顺便说一句,我是这里的新手,请不要欺负:)
EDIT: I tried changing $got_users['nombre'];
编辑:我试图更改
$got_users['nombre'];
with $searched_for
to see if $searched_for
is empty and yes it doesn't return any string that's why I am getting all the rows. 用
$searched_for
查看$searched_for
是否为空,是的,它不返回任何字符串,这就是为什么我要获取所有行的原因。 $query
is getting an empty variable but Why? $query
得到一个空变量,但是为什么呢?
Here's my HTML: 这是我的HTML:
<form target="u-n" id="search_input" action="search_user.php" method="post">
<input id="search-input" name="searched_for" type="search" placeholder="Search">
</form>
You used <input type="search" />
which is a HTML5 feature. 您使用了HTML5功能的
<input type="search" />
。 Older browsers may not support this. 较旧的浏览器可能不支持此功能。 Replace this input with
type="text"
. 将此输入替换为
type="text"
。
Then, your $_POST['searched_for']
should populate properly, that is: 然后,您的
$_POST['searched_for']
应该正确填充,即:
<input name="searched_for" type="text" placeholder="Search" />
Also, you used the same id
multiple times, which is an invalid HTML syntax. 另外,您多次使用相同的
id
,这是无效的HTML语法。
Reference: HTML input tag at MDN 参考: MDN上的HTML输入标签
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