duancheng3042 2016-02-26 18:36
浏览 48

php jquery显示来自mysql getjson的数据

I need to show data retrieved from mysql in a page. My codes are :

jquery:

$('#search').on('click', function() { 
        alert($('.search_word').val());
        var data_to_php = $('.search_name:visible').val(); //take the text only from the visible search box
        alert(data_to_php);


        //////////////////////////////////////////////////////////////////////
        //The following code will get the search list populated  by //////////
        //retrieving data from mysql table////////////////////////////////////
        //////////////////////////////////////////////////////////////////////

        $.getJSON('./search.php', 'keyword='+data_to_php, function(info) {

                $.each(info, function() {
                    $('ul.search_list').append('<li   class="search_data">hi</li>'); //this code is for the purpose of testing only

                    console.log(info);
                }); 


        });

    }); //end #search.on.click

the php code is :

require(connect.php);

$keywords = $_GET['keyword'];

$query = "SELECT name FROM test ORDER BY name WHERE name LIKE   '%".$_GET['keyword']."%'";

$result = mysqli_query($conn,$query); //connection variable
                                      //$conn is from
                                      //included file
                                      //connect.php 

$jsonData = array();

while($row = mysqli_fetch_array($result)){

    $jsonData[] = $row;

}

echo json_encode($jsonData);
?>

In console.log output, I am getting :

GET ./search.php?keyword=s 500 Internal Server Error with blank response. Internal error 500 can as well be generated if php is not working properly. There might also be some error in the code.

Now phpinfo is showing results properly.

How do I know if the getjson call to php script is successful?

  • 写回答

1条回答 默认 最新

  • dongmei1988 2016-02-26 18:43
    关注

    There is an error in your MySQL query, you cannot call ORDER BY before WHERE clause, modify your query,

    SELECT name FROM test WHERE name LIKE '%".$_GET['keyword']."%' ORDER BY name

    A very good reference to know the correct order of execution in MySQL - MySQL query / clause execution order

    Also, your file name in the require function should be enclosed within quotes - require('connect.php');

    评论

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