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Jun 23, 2016 pm 01:08 PM

我是想实现刚刚开始显示所有数据库表的信息，按照输入框查询后，只显示查询到的内容，其余的不出现，怎么改呢，我现在只做到了显示全部信息！！
include("inc/conn.php");
?>

测试








        if(!isset($_POST['search'])||trim($_POST['search'])==''){
        $_POST['search']='';
    }
    $find = $_POST['search'];
$result=mysql_query("select * from job where title like '%".$inputkey."%'");
   while($myrow=mysql_fetch_array($result)){
   ?>

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回复讨论(解决方案)

表单写错了

改成这样

$inputkey 这个变量是从哪里获取？

你这一行里面没有name 指，php接受的表单值是根据name值接受的。
我们公司的前端写页面的时候也总是这样不写name值。。。。。。

input 不加name属性，php怎么获取得到？

取值查询也有问题
$inputkey = isset($_POST['inputkey ']) ? $_POST['inputkey '] : '';
$result=mysql_query("select * from job where title like '%".$inputkey."%'");

谢谢各位，问题解决了，最后代码是这样的
include("inc/conn.php");
?>

测试








    //    if(!isset($_POST['search'])||trim($_POST['search'])==''){
//        $_POST['search']='';
//    }
//    $find = $_POST['search'];
$inputkey = isset($_POST['inputkey']) ? $_POST['inputkey'] : '';
    $result=mysql_query("select * from job where title like '%".$inputkey."%'");
   while($myrow=mysql_fetch_array($result)){
   ?>

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