使用PHP獲取Mysqli數據庫搜索的URL參數

Question

我想了解如何從PHP獲取URL參數。 所以我的代碼有點阿賈克斯開始射擊的PHP和警報與反應：

$.ajax({          
    url: 'blog/assets/php/load-blog.php', 
    dataType: 'json',  
    success: function(data){ 
    alert(data); 
    } 
    }); 

然後，我有我的PHP：

<?php 
    //==== CONNECTION VARIABLE 
    $con = new mysqli($host,$user,$pass,$database); 


    //==== GET URL PARAMETER 
    $urlParam = $_GET["date"]; 


    //==== FETCH DATA 
    $result = $con->query("SELECT * FROM $table WHERE DATE = '$urlParam'"); 


    //==== CREATE ARRAY 
    $blogArray = array(); 
    while ($row = $result->fetch_row()) { 
    $blogArray[] = $row; 
    } 


    //==== ECHO AS JSON 
    echo json_encode($blogArray); 
?> 

的URL會是這個樣子：

http://www.mysite.com/blog/?date=2013-05-14 

所有這一切的問題是沒有正確選擇URL參數，但我不知道我做錯了什麼。

Answer 1

的Javascript

var date = '2013-05-14'; 

$.ajax({   
    type: "GET",        
    url: 'blog/assets/php/load-blog.php', 
    dataType: 'json', 
    data: { date: date } 
    success: function(data){ 
    alert(data); 
    } 
}); 

PHP

<?php 
//==== CONNECTION VARIABLE 
$con = new mysqli($host,$user,$pass,$database); 


//==== GET URL PARAMETER 
$urlParam = $_GET["date"]; 

//==== PREPARED STATEMENT 
$stmt = $con->prepare("SELECT id,name FROM __TABLE_NAME__ WHERE date = ?"); 
$stmt->bind_param("s", $urlParam); 
$stmt->execute(); 
$stmt->bind_result($id, $name); 

//==== CREATE ARRAY 
$blogArray = array(); 

//==== FETCH DATA 
while ($stmt->fetch()) { 
    $blogArray[] = array(
    'id' => $id, 
    'name' => $name 
    ); 
} 

//==== CLOSE STATEMENT 
$stmt->close(); 

//==== ECHO AS JSON 
echo json_encode($blogArray); 
?>