PHP SQL更新查詢無法正常工作

Question

我一直在嘗試創建一個簡單的表單來更新數據庫的詳細信息，PHP代碼如下：

// UPDATE THE DATABASE RECORDS //
$update = $_GET['update'];
if($update == "true"){
    $setDetails="UPDATE users SET email='{$_POST['email']}', api_key='{$_POST['api_key']}', api_secret='{$_POST['api_secret']}' WHERE username='{$_POST['username']}'";
    if(mysql_query($setDetails)){
        $updatemsg = '<div class="alert alert-success"><a href="#" class="close" data-dismiss="alert">×</a><strong>Success!</strong> Your details have been updated in our database.</div>';
    }else{
        $updatemsg = '<div class="alert alert-error"><a href="#" class="close" data-dismiss="alert">×</a><strong>Failure!</strong> Your details could not be updated in our database. Please try again later or contact us if this keeps happening.</div>';
    }
}else if($update == "false"){
    $updatemsg = '<div class="alert alert-success"><a href="#" class="close" data-dismiss="alert">×</a><strong>Success!</strong> Your changed were discarded.</div>';
}
// UPDATE THE DATABASE RECORDS //

// GET THE DATABASE RECORDS //
$getDetails="SELECT * FROM users WHERE username='$username'";
$details=mysql_query($getDetails);
$num=mysql_numrows($details);
if($num != 0){
    $new_user = false;
    $username=mysql_result($details,0,"username");
    $email=mysql_result($details,0,"email");
    $subscription_type=mysql_result($details,0,"subscription_type");
    $subscription_date=mysql_result($details,0,"subscription_date");
    $api_key=mysql_result($details,0,"api_key");
    $api_secret=mysql_result($details,0,"api_secret");
    setcookie("api_key", $api_key, time()+50000);
    setcookie("api_secret", $api_secret, time()+50000);
}else{
    $new_user = true;
}
// GET THE DATABASE RECORDS //

獲取數據庫記錄時定義的變量隨后用於填充HTML表單：

<form action="?update=true" method="POST">
<h2>Your Details</h2>
<input id="username" name="username" type="text" placeholder="" disabled="true" class="input-xlarge" value="<?=$username?>">
<input id="email" name="email" type="text" placeholder="" class="input-xlarge" value="<?=$email?>">
<input id="subscription_type" name="subscription_type" type="text" placeholder="" disabled="true" class="input-xlarge" value="<?=$subscription_type?>">
<input id="subscription_date" name="subscription_date" type="text" placeholder="" disabled="true" class="input-xlarge" value="<?=$subscription_date?>">
<input id="api_key" name="api_key" type="text" placeholder="" class="input-xlarge" value="<?=$api_key?>">
<input id="api_secret" name="api_secret" type="text" placeholder="" class="input-xlarge" value="<?=$api_secret?>">
<button type="submit" class="btn btn-success" id="saveChanges"><i class="icon-ok icon-white"></i> Save Changes</button> <a href="?update=false" class="btn btn-danger" id="discardChanges"><i class="icon-remove icon-white"></i> Discard Changes</a>
</form>

當第一次加載頁面時，表單填充沒有問題，但是當它被編輯並提交時， $updatemsg是第二個（成功），但數據庫中的數據沒有變化。 有任何想法嗎？

Answer 1

看起來它失敗了因為引號和括號被錯誤地使用了。 但你需要做的是：

$email = mysql_real_escape_string($_POST['email']);

....

"UPDATE users SET email='$email'.....

等等，等等。 如果不是這樣，在else部分調用mysql_error（），它會告訴你查詢是否失敗。

Answer 2

首先，我會在一個常量文件中聲明api機密，而不是通過頁面上的http帖子。 嘗試以下操作以使更新生效，假設帖子值存在

$ setDetails =“UPDATE users SET email ='”。$ _ POST ['email']。“'，api_key ='”。$ _ POST ['api_key']。“'，api_secret ='”。$ _ POST ['api_secret' '。'WHERE username ='“。$ _ POST ['username']。”'“;