javascript - Ajax:500 内部服务器错误

Question

我正在使用 ajax 从我的数据库中获取事件。检索结果不起作用，没有任何显示，并且在控制台中我收到以下错误消息:

POST http://www.example.com/system/live_filter.php 500 (Internal Server Error) jquery.min.js:4

这是我的HTML/JS:

<div id="results">

<script type="text/javascript">

    // 1. When user comes on page from homepage, results will be fetched with ajax
    function updateResults() { 

        // 2. Create array with values of all filter fields
        var value_town_id = $('#town_id').val();
        var value_type = $('#filter_type').val();
        var value_date = $('#filter_date').val();
        var array_filter_values = new Array(value_town_id, value_type, value_date);
        array_filter_values.join(', ');
        query_value = array_filter_values;

    // 3. Start ajax
    $.ajax({
            type: "POST",
            url: "system/live_filter.php",
            data: { query: query_value },
            cache: false,
            success: function(html){
                $("#results").html(html);
            }
        });

    };

    // 4. FIRE FUNCTION!
    updateResults();

</script>

</div>

这是我的 live_filter.php，值通过 Ajax 发送到它:

require_once 'db.php';

// Define Output HTML Formating
$html = '';
$html .= '<div class="event">';
$html .= '<h3>titleString</h3>';
$html .= '<p>typeString</p>';
$html .= '<p>dateString</p>';
$html .= '</div>';

// Get values
$values_string = $_POST['query'];

// Explode to array
$values_array = explode(',', $values_string);
$town_id = $values_array[0];
$type = $values_array[1];
$date = $values_array[2];

// Prepare values for database results query
$town_id = $db->real_escape_string($town_id);
$type = $db->real_escape_string($type);
$date = $db->real_escape_string($date);


// Build Query
$query = "SELECT * FROM events WHERE towns_id=$town_id AND type='$type' AND date>=$date";

// Do Search
$results = $db->query($query);
while($result = $results->fetch_assoc()) {

            // Format Output Strings And Hightlight Matches
            $display_title = $result['title'];
            $display_type = $result['type'];
            $display_date = $result['date'];

            // Insert title
            $output = str_replace('titleString', $display_title, $html);
            // Insert type
            $output = str_replace('typeString', $display_type, $output);
            // Insert date
            $output = str_replace('dateString', $display_date, $output);

            // Output
            echo($output);
}

有人知道问题出在哪里吗？

我收到以下错误消息:您的 SQL 语法有误；查看与您的 MySQL 服务器版本对应的手册，了解在第 1 行的“AND type='' AND date>=''' 附近使用的正确语法

Answer 1

如果变量$town_id 为空 那么只有sql 产生错误，你能打印Post 值吗

SELECT * FROM events WHERE towns_id = AND type='$type' AND date>='$date'

如果你想执行 towns_id 为空的查询。将所有变量附加​​到 single

SELECT * FROM events WHERE towns_id = '$town_id' AND type='$type' AND date>='$date'