用PHP更新MySQL數據庫

Question

基本上我正在嘗試使用PHP來更新MySQL數據庫，並且正在用HTML表單測試它。

我打算在android應用程序中使用這個值，所以這些值是從哪裏獲取的，但是目前我只是用HTML表單測試來測試PHP代碼。當我使用HTML表單進行測試時，當前沒有更新適當的數據。

我的代碼導致這種情況有什麼問題？

PHP代碼：

/* 
* Following code will create a new product row 
* All player details are read from HTTP Post Request 
*/ 

// array for JSON response 
$response = array(); 

// check for required fields 
if (isset($_POST['PlayerID']) && isset($_POST['Score']) && isset($_POST['LastHolePlayed'])&&  
isset($_POST['Overall'])) { 

$playerid = $_POST['PlayerID']; 
$score = $_POST['Score']; 
$lastholeplayed = $_POST['LastHolePlayed']; 
$overall = $_POST['Overall']; 

// include db connect class 
require('db_connection.php'); 



// mysql inserting a new row 
$result = mysql_query("UPDATE `week1` SET Score = `$score`, LastHolePlayed = `$lastholeplayed`, 

Overall` = $overall` WHERE PlayerID = `$playerid`"); 


// check if row inserted or not 
if ($result) { 
    // successfully inserted into database 
    $response["success"] = 1; 
    $response["message"] = "Player successfully added."; 

    // echoing JSON response 
    echo json_encode($response); 
} else { 
    // failed to insert row 
    $response["success"] = 0; 
    $response["message"] = "An error occurred."; 

    // echoing JSON response 
    echo json_encode($response); 
} 
} else { 
// required field is missing 
$response["success"] = 0; 
$response["message"] = "Required field(s) is missing"; 

// echoing JSON response 
echo json_encode($response); 
} 

的html代碼：

<form action="http://localhost/realdeal/updateplayer.php" method="POST"> 
PlayerID <input type="text" id='PlayerID' name='PlayerID'><br/><br/> 
Score <input type="text" id='Score' name='Score'><br/><br/> 
LastHolePlayed <input type="text" id='LastHolePlayed' name='LastHolePlayed'><br/><br/> 
Overall <input type="text" id='Overall' name='Overall'><br/><br/> 

    &nbsp; <input type="submit" value="submit"> 

</form> 

Answer 1

您所查詢的分隔符，需要加以糾正：

$result = mysql_query("UPDATE `week1` SET Score = '$score', `LastHolePlayed` = '$lastholeplayed', `Overall` = '$overall' WHERE `PlayerID` = '$playerid'"); 

注意列周圍的反引號（`）和圍繞值的單引號（'）。

此外，當您正在調試的查詢時，總是檢查MySQL的錯誤：

$result mysql_query(...) or die("Query failed: " . mysql_error()); 

最後，你應該知道你的查詢讓你對SQL注入攻擊開放。在將其包含在查詢中之前，始終清理輸入數據。

Answer 2

改變您的查詢：

$result = mysql_query("UPDATE `week1` SET `Score` = '$score', `LastHolePlayed` = '$lastholeplayed', `Overall` = '$overall' WHERE `PlayerID` = '$playerid'"); 

Answer 3

你的sql語句是錯誤的。你可以按照上面的描述寫，或者你可以直接寫下沒有任何撇號符號的語句 - $ result = mysql_query（「UPDATE week1 SET Score = $ score，LastHolePlayed = $ lastholeplayed，Overall = $ WHERE PlayerID = $ playerid」） ;

此外，您能否解釋「適當的數據未被更新」是什麼意思。如果你給出/陳述你得到的錯誤，那將會更加清楚。