php搜索函数使用javascript动态显示结果 - php search function dynamically show results with javascript

php search function dynamically show results with javascript

我有以下php脚本，它运行良好，它使用搜索词并将其与几个不同的字段进行比较，然后打印出匹配的每个记录：

<?php
mysql_connect ("localhost", "root","")  or die (mysql_error());  
mysql_select_db ("table"); 
$search = isset($_POST['search']) ? $_POST['search'] : '';
$sql = mysql_query("select * from asset where 
                             name like '%$search%' or 
                             barcode like '%$search%' or
                             serial like '%$search%' ");  
while ($row = mysql_fetch_array($sql)){  
    echo '<br/> Name: '.$row['name'];
    echo '<br/> Barcode: '.$row['barcode'];
    echo '<br/> Serial: '.$row['serial']; 
    }
?>

这就是链接到它的表单：

<form action="http://localhost/test/search.php" method="post">
 Search: <input type="text" name="search" /><br />
<input type="submit" name="submit" value="Submit" />
</form>

我需要了解如何对搜索结果进行编码，这样我就可以在javascript函数中使用它们，然后我就可以将它们显示在表单下面的同一个html页面上。

为此，您必须使用AJAX。您可以使用JSON将数据发送回同一页面。
建议-不要使用mysql_*函数，因为它们已被弃用。学习mysqli_*并尝试使用它。

<script>
$(function(ev){
   ev.preventDefault();
   $("form").on('submit', function(){
       var form = $(this);
       var url = form.attr('action');
       var data = form.serialize();
       $.post(url, data)
       .done(function(response){
            if(response.success == TRUE)
            {
                  // Search result found from json
                  // You have to loop through response.data to display it in your page
                  // Your single loop will have something like below -
                  var name = response.data.name;
                  var barcode = response.data.barcode;
                  var serial = response.data.serial;
                  $("#name").html(name);
                  $("#barcode").html(barcode);
                  $("#serial").html(serial);
            }
            else
            {
                  // search result not found
            }
       });
   });
});
</script>

打开search.php

<?php
mysql_connect ("localhost", "root","")  or die (mysql_error());
mysql_select_db ("table");
$search = isset($_POST['search']) ? $_POST['search'] : '';
$sql = mysql_query("select * from asset where 
                         name like '%$search%' or 
                         barcode like '%$search%' or
                         serial like '%$search%' ");
$num = mysql_rows_nums($sql);
$json = array();
if($num > 0)
{
    $json['success'] = TRUE;
    while ($row = mysql_fetch_array($sql)){  
      $json['data']['name'] = $row['name'];
      $json['data']['barcode'] = $row['barcode'];
      $json['data']['serial'] = $row['serial'];
    }
}
else
{
    $json['success'] = FALSE;
}
return json_encode($json);

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