Question

我想在点击任何搜索结果后自动填写id = id，name，email，company ..的div。搜索结果中的id用作过滤器，以从Mysql中获取相应的行数据来自search.php

中使用的同一个表

这是我的表单

<link href="../action/css/onlinecustom.css" rel="stylesheet"     type="text/css">
<script src="http://code.jquery.com/jquery-1.10.2.js"  type="text/javascript"></script>
<script src="./action/scripts/global2.js" type="text/javascript"></script> 

<script>
function searchq() {
    var searchTxt = $("input[name='search']").val();

     $.post("../action/subs/search.php/", {searchVal: searchTxt}, function(output) {
         $("#output").html(output);
     });
 }

</script>

<title>Search</title>

<body>
<form action="http://comiut.com/index.php/user-records" method="post">
  <input type="text" name="search" Placeholder="enter the search    criteria..." onkeydown="searchq();"/>
  <input type="submit" value ="serach"/>

</form>
//Serach result//
<div id="output"> </div>

//Data to populate upon click on any search result//
    <div id="id"></div> 
    <div id="name"></div>
    <div id="email"></div>
    <div id="company_name"></div>

</body>

**我创建了一个global2.js文件**

jQuery('body').on('click', 'a.resultItem', function(e) {
e.preventDefault();
jQuery.ajax({
    url: "http://onlinepcdoc.com/action/subs/getItem.php",
    method: 'post',
    data : jQuery(this).data('id') // see the data attribute we used above in the a tag we constructed
}).done(function(data) {
    jQuery("#id").html(data.id);
    jQuery("#name").html(data.name);
    jQuery("#email").html(data.email);
    jQuery("#company_name").html(data.company);
   });
});

我还创建了search.php

<?php 

    include '../db/connect6.php';


if(isset($_POST['searchVal'])) {
  $searchq = $_POST['searchVal'];
  $searchq = preg_replace ("#[^0-9a-z]#i","",$searchq);

  $query = mysql_query("SELECT * FROM oz2ts_users WHERE oz2ts_users.id LIKE '%$searchq%' OR oz2ts_users.name LIKE '%$searchq%'") or die("Could not search"); 
  $count = mysql_num_rows($query);
  if($count == 0){
     $output = 'There is no result to show!'; 
       } else{ 
        while($row = mysql_fetch_array ($query)) {
            $id = $row['id'];
            $name = $row['name'];
            $username = $row['username'];    

      $output .= '<div><a class="resultItem" data-id="' . $id . '">'   
       . $name . ' '.$username.'</a></div>';   

   }            
 }

 }
echo($output);
?>

**这是getItem.php **

<?php

include '../db/connect6.php';

if(isset($_POST['id'])) {
    $id = intval($_POST['id']);
    $result = mysqli_query("SELECT oz2ts_users.id, oz2ts_users.name,    oz2ts_users.username,  oz2ts_users.email FROM oz2ts_users WHERE oz2ts_users.id =      $id") or die("Could not search"); 
    // since we expect only one result we don't need a loop
    $row = mysqli_fetch_assoc($result);
    // let's return the $row in json format
    // first let's prepare the http header
    header('Content-Type: application/json');
    // and now we return the json payload
    echo json_encode($row);
}

?>

Answer 1

你走在正确的道路上，虽然你自己没有尝试过任何东西，但这里有一些东西让你继续前进：

您应该返回类似的内容并填充搜索结果列表

$output .= '<div><a class="resultItem" data-id="' . $id . '">'   
           . $name . ' '.$username.'</a></div>';

接下来你需要一个额外的ajax调用来填充底部信息所以这里是javascript位，注意使用click {{1 body事件绑定到动态创建元素的方式构造：

on

现在你只需要$('body').on('click', 'a.resultItem', function(e) { e.preventDefault(); $.ajax({ url: "getItem.php", method: 'post', data : $(this).data('id') // see the data attribute we used above in the a tag we constructed }).done(function(data) { $("#id").html(data.id); $("#name").html(data.name); $("#email").html(data.email); $("#company_name").html(data.company); }); });就可以这样：

getItem.php

搜索结果的动态链接

1 个答案: