Question

您好，我正在尝试使用我的ajax php向我的网站添加实时搜索功能

我希望在单击实时搜索结果时更改实时搜索字段的值并将所选结果的ID放在隐藏的表单字段中，以便稍后插入

我尝试在下面的代码中执行此操作，但出现以下错误：参数列表后未捕获到的SyntaxError：缺少

<script>
    $(document).ready(function(){
        load_data();
        function load_data(query)
        {
            $.ajax({
                url:"search.php",
                method:"GET",
                data:{textbook:query},
                success:function(data)
                {
                    $('#result').html(data);
                }
            });
        }
        $('#search').keyup(function(){
            var search = $(this).val();
            if(search != '')
            {
                load_data(search);
            }
            else
            {
                load_data();
            }
        });
    });
</script>

<script>

    selectxt(id, textbood_adress){

        $('#search').val(textbood_adress);

    }
</script>

<div class="form-group input-group" id="textbook">
<input type="text" name="search" id="search" placeholder="Search"  class="form-control">
<span class="input-group-btn">
<button class="btn btn-default" type="button"><i class="fa fa-search"></i>
</button>
</span>

</div>
<div id="result"></div>

和php

include ("../../includes/config.php");
$output = '';

if(isset($_GET['textbook'])){

    $key=$_GET['textbook'];
$key = $db->escape($key);



    $results = $db->rawQuery("SELECT * from textbook where textbook_address like '%{$key}%' ");

    if($db->count > 0){

        $output .= '
      <div class="table-responsive">
       <table class="table table bordered">
       <tr>
         <th>Textbook address</th>
         <th>select?</th>
        </tr>';

        foreach ($results as $result) {

            $output .= '
  <tr>
    <td>'.$result["textbook_address"].'</td>
    <td><a id="selectclick" href="#" onclick="selectxt('.$result['id'].','.$result['textbook_address'].')">select</a></td>
   </tr>
  ';


        }

        echo $output;

    }else{

        echo "<span>No results for your search</span>";

    }






}



?>

Answer 1

您的函数声明错误。更改您的JavaScript代码来自

<script>
$(document).ready(function(){
    load_data();
    function load_data(query)
    {
        $.ajax({
            url:"search.php",
            method:"GET",
            data:{textbook:query},
            success:function(data)
            {
                $('#result').html(data);
            }
        });
    }
    $('#search').keyup(function(){
        var search = $(this).val();
        if(search != '')
        {
            load_data(search);
        }
        else
        {
            load_data();
        }
    });
});
</script>
<script>

selectxt(id, textbood_adress){

    $('#search').val(textbood_adress);

}
</script>

到

<script>

function load_data(query) {
    $.ajax({
        url: "search.php",
        method: "GET",
        data: {textbook: query},
        success: function (data) {
            $('#result').html(data);
        }
    });
}

function selectxt(id, textbood_adress) {
    $('#search').val(textbood_adress);
}

$(document).ready(function () {
    load_data();
    $('#search').keyup(function () {
        var search = $(this).val();
        if (search != '') {
           load_data(search);
        }
        else {
           load_data();
        }
     });
});
</script>

您的javascript函数声明有误，也放错了位置。这是一个示例on CodePen。

请检查并确保$result['textbook_address']中没有'。这会破坏您的html输出，因为浏览器会尝试将其解释为错误

从ajax php脚本获取数据

1 个答案: