PHP搜索过滤函数不使用';行不通
我在php表单中创建了搜索过滤器函数。从表中的mysql获取数据。所有的查询和其他都写得很好。但搜索功能不起作用PHP搜索过滤函数不使用';行不通,php,mysql,mysqli,Php,Mysql,Mysqli,我在php表单中创建了搜索过滤器函数。从表中的mysql获取数据。所有的查询和其他都写得很好。但搜索功能不起作用 <?php require_once 'init.php'; include 'header.php'; include 'navigation.php'; if(isset($_POST['search'])){ $valSearch = $_POST['searchButton']; $query = "SELECT
<?php
require_once 'init.php';
include 'header.php';
include 'navigation.php';
if(isset($_POST['search'])){
$valSearch = $_POST['searchButton'];
$query = "SELECT * FROM mainTable WHERE CONCAT ('id','commandName','commandDesc','status') LIKE '%".$valSearch."%'";
$featured = filterTable($query);
}else{
$query = "SELECT * FROM mainTable";
$featured = filterTable($query);
}
function filterTable($query){
$connect = mysqli_connect("localhost","root","","linuxcommanddictionary");
$filter_result = mysqli_query($connect,$query);
return $filter_result;
}
?>
<form action = "index.php" method="post">
<div class="container">
<h3>Linux Fedora OS</h3><hr>
<button type="submit" name ="searchButton" class="btn btn-default"><span class="glyphicon glyphicon-search"></span></button>
<input type="text" name = "search" class="form-control" placeholder="Хайх үгээ бичнэ үү">
<div class="col-md-12">
<div class="row">
<table class="table table-bordered">
<thead>
<th>ID</th>
<th>Command Name</th>
<th>Command Description</th>
<th>Status</th>
</thead>
<tbody>
<?php while($result = mysqli_fetch_assoc($featured)): ?>
<tr>
<td><?=$result['id']; ?></td>
<td><?=$result['commandName']; ?></td>
<td><?=$result['commandDesc']; ?></td>
<td><?=$result['status']; ?></td>
</tr>
<?php endwhile; ?>
</tbody>
</table>
</div>
</div>
</div>
</form>
<?php
include 'footer.php';
?>
</body>
</html>
如注释中所述,您在函数内调用mysqli连接,然后在函数的“作用域”外请求mysqli_fetch_assoc()
在全局作用域中创建连接器,并在函数中调用“global”以获取数据(cpu吸盘)-或-将所有内容移动到全局作用域中的其他位置。既然您说没有任何错误,请尝试以下操作: 1-通过phpmyadmin手动运行查询,查看是否得到任何结果 2-将连接变量作为全局变量传递。阅读更多关于变量的内容,以更好地理解所涉及的概念
function filterTable($query){
global $connect;
$connect = mysqli_connect("localhost","root","","linuxcommanddictionary");
$filter_result = mysqli_query($connect,$query);
// fetch results here as mysqli_query returns true/false.
$final_results = array();
while($row=mysqli_fetch_array($filter_result,MYSQLI_NUM)){
$final_results[] = $row;
}
return $final_results;
}
3-打开错误报告以查看是否出现任何问题开发人员对此并不是很好的解释。您会遇到什么错误?另外,您似乎需要在筛选表中将连接变量作为全局变量传递。没有出现任何错误。另外:$valSearch=$\u POST['searchButton'];然后回应它,它说什么?那么:$valSearch=$_POST['search'];呢?你的问题太少了,我甚至试着编辑它,你甚至没有足够的“内容”可以编辑。请做你的家庭作业,这样我们才能更好地帮助你。谢谢明白:)
function filterTable($query){
global $connect;
$connect = mysqli_connect("localhost","root","","linuxcommanddictionary");
$filter_result = mysqli_query($connect,$query);
// fetch results here as mysqli_query returns true/false.
$final_results = array();
while($row=mysqli_fetch_array($filter_result,MYSQLI_NUM)){
$final_results[] = $row;
}
return $final_results;
}