jQuery UI自动完成搜索结果不显示

Angie 发表于 Dev

安吉

我无法使用php从mysql数据库获得搜索结果。

PHP代码：

require_once "connectmysql.php";
$belongsto=$current_user->businessname;
$q = trim(strip_tags($_GET["term"]));
if (!$q) return;
$sql = "select clientname as value from zb_clients where clientname LIKE '%".$q."%' AND belongsto='".$belongsto."'";
$result = mysql_query($sql);
while($row = mysql_fetch_array($result,MYSQL_ASSOC)) {
    $row['value']=htmlentities(stripslashes($row['value']));
    $row_set[] = $row;
}
echo json_encode($row_set);

jQuery代码：

<link rel="stylesheet" href="http://ajax.googleapis.com/ajax/libs/jqueryui/1.10.1/themes/base/minified/jquery-ui.min.css" type="text/css" />
<script type="text/javascript" src="http://code.jquery.com/jquery-1.9.1.min.js"></script>
<script type="text/javascript" src="http://code.jquery.com/ui/1.10.1/jquery-ui.min.js"></script>

<script type="text/javascript">
$(function() {

    //autocomplete
    $("#search").autocomplete({
        source: "../searchclient.php",
        minLength: 1,
    });             

});
</script>

输入栏位：

<input type="text" name="search" id="search" placeholder="Search for Business Name" />

我相信php代码是正确的。如果我自己运行php代码并使用/searchclient.php?term=a

例如，它在数组中返回我想要的结果。

例如[{"value":"Hello World"},{"value":"East Meets West JV"}]。

如果我替换了Jquery行

source: "../searchclient.php",

和

source: [ "c++", "java", "php", "coldfusion", "javascript", "asp", "ruby" ],

然后自动完成功能将与该来源一起使用。因此，将数组传递回JQuery必须存在一个问题。

我不能完全把手指放在上面。我是否缺少一些重要的东西？

香港专业教育学院试图调试与萤火虫，但它不返回任何错误。

任何帮助将不胜感激！

编辑的PHP代码：

require_once "connectmysql.php";
$belongsto=$current_user->businessname;
$q = $_GET["term"];
if (!$q) return;
$sql = "select clientname as value, idzb_clients as id from zb_clients where clientname LIKE '%".$q."%' AND belongsto='".$belongsto."'";
$result = mysql_query($sql);
while($row = mysql_fetch_array($result,MYSQL_ASSOC)) {
    $row['id']=htmlentities(stripslashes($row['id']));
    $row['value']=htmlentities(stripslashes($row['value']));
    $row['label']=htmlentities(stripslashes($row['value']));
    $row_set[] = $row;
}
echo json_encode($row_set);

安吉

我沿着AJAX路线走了，这似乎行得通：

的HTML

$( "#search" ).autocomplete({
  source: function( request, response ) {
    $.ajax({
      url: "../searchclient.php",
      dataType: "json",
      data: {
        q: request.term
      },
      success: function( data ) {
        response( data );
      }
    });
  },
  minLength: 1
});

SEARCHCLIENT.PHP

<?php

require_once "connectmysql.php";
$belongsto = $current_user->businessname;
$q = $_GET['q'];
$sql = "select clientname as value, idzb_clients as id, has_ledger_setup from zb_clients where clientname LIKE '%".$q."%' AND belongsto='".$belongsto."'";
$result = mysql_query($sql);
while($row = mysql_fetch_array($result,MYSQL_ASSOC)) {
    $row['id'] = htmlentities(stripslashes($row['id']));
    $row['value'] = htmlentities(stripslashes($row['value']));
    $row['label'] = htmlentities(stripslashes($row['value']));
    $row_set[] = $row;
}
header('Content-Type: application/json');
echo json_encode($row_set);

?>