Question

我想为我的房地产网站创建搜索功能。我已创建搜索条件并满足此条件我使用了jax调用。这个ajax将调用php文件，该文件将数据发送到搜索页面。这是我的示例代码。我不知道那里有什么问题。谢谢你

  function searchBuy() {
        //alert("search Loaed");
       $.ajax({
        url: 'searchBuyCriteria.php',
        type: 'POST',
        //dataType: "json",
        data: {
            cat: $('#SelectCat').val(),
           MaxArea:$('#SelectMaxArea').val()
        }
        }).success(function(data){
            console.log("Success"+data);
            alert(JSON.stringify(data));
        }).error(function(data){
            console.log("Error"+data);
        });
        } </scipt>

searchCriteria.php

    <?php

    include_once('_secure/conn.php');

    $city = $_POST['SelectCity'];
    $cat= $_POST['SelectCat'];
    $rooms = $_POST['SelectRooms'];
    $minRange = $_POST['SelectMinRange'];
    $maxRange = $_POST['SelectMaxRange'];
    $minArea = $_POST['SelectMinArea'];
    $maxArea = $_POST['SelectMaxArea'];

    $conn=mysql_connect(DB_HOST,DB_USER,DB_PWD,DB_NAME);
    mysql_select_db(DB_NAME);

    $Searchqry ="Select * from tblcategory where category_title like '%".$cat."%' ";
    $searchExec = mysql_query($Searchqry) or die(mysql_error());
    $searchRow = mysql_fetch_array($searchExec);
    //echo $searchRow;
    echo json_encode($searchRow);
?>

conn.php

<?php 
session_start();    
ob_start();
    if(!isset($path))
        $path="./";
    elseif($path=="")
        $path="../";    

    require_once("settings.inc.php");
    $database='srijanip';
    $user='srijanip';
    $pass='********';
    $host='example.*****.in';

    define("DB_HOST", "$host");
    define("DB_NAME", "$database");
    define("DB_USER", "$user");
    define("DB_PWD", "$pass");
?>

警告：session_start（）[function.session-start]：无法发送会话cookie -

0 个答案: