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我是一個完整的代碼nooby和幾個月前建立一個基本的網站之前，我的服務器主板死亡。我只是重建了服務器並複製了我備份到桌面的所有網站/數據庫代碼。據我所知，php代碼和數據庫設置之間的一切都是一樣的。PHP數據庫入口問題（Wampserver）
我的網站工作90％，除了當我去輸入數據。與預期的「成功創建新記錄」相反，我收到了「等待輸入」的錯誤。以下代碼的最底層上的相關if語句爲。
有沒有人偶然有任何線索爲什麼輸入我的所有正確的數據後，它不會觸發「成功創建新記錄」？如果這意味着任何代碼工作正常，我的桌面上，除非我做了一件很愚蠢的我適當/ pasta'd一切複製到該服務器:(
<?php
//Connect to database
include('database.php');
//grab user logged in for database insertion
$user_check = $_SESSION['login_user'];
$ses_sql = mysqli_query($dbcon,"select username from t_login where username = '$user_check' ");
$row = mysqli_fetch_array($ses_sql,MYSQLI_ASSOC);
$login_session = $row['username'];
// Create DB Variable
$login_session = $row['username'];
$excercisename = (isset($_POST['excercisename']) ? $_POST['excercisename'] : null);
$setcount = (isset($_POST['setcount']) ? $_POST['setcount'] : null);
$rep1 = (isset($_POST['rep1']) ? $_POST['rep1'] : null);
$rep2 = (isset($_POST['rep2']) ? $_POST['rep2'] : null);
$rep3 = (isset($_POST['rep3']) ? $_POST['rep3'] : null);
$rep4 = (isset($_POST['rep4']) ? $_POST['rep4'] : null);
$rep5 = (isset($_POST['rep5']) ? $_POST['rep5'] : null);
$rep6 = (isset($_POST['rep6']) ? $_POST['rep6'] : null);
$rep7 = (isset($_POST['rep7']) ? $_POST['rep7'] : null);
$rep8 = (isset($_POST['rep8']) ? $_POST['rep8'] : null);
$weight1 = (isset($_POST['number1']) ? $_POST['number1'] : null);
$weight2 = (isset($_POST['number2']) ? $_POST['number2'] : null);
$weight3 = (isset($_POST['number3']) ? $_POST['number3'] : null);
$weight4 = (isset($_POST['number4']) ? $_POST['number4'] : null);
$weight5 = (isset($_POST['number5']) ? $_POST['number5'] : null);
$weight6 = (isset($_POST['number6']) ? $_POST['number6'] : null);
$weight7 = (isset($_POST['number7']) ? $_POST['number7'] : null);
$weight8 = (isset($_POST['number8']) ? $_POST['number8'] : null);
$cardio = (isset($_POST['cardio']) ? $_POST['cardio'] : null);
$duration = (isset($_POST['duration']) ? $_POST['duration'] : null);
$weight = (isset($_POST['weight']) ? $_POST['weight'] : null);
$comments = (isset($_POST['comments']) ? $_POST['comments'] : null);
//Execute the insert statement
if(!empty($excercisename)){
$sql = "INSERT INTO t_workout (username, excercisename, setcount, set1reps, set2reps, set3reps, set4reps, set5reps, set6reps, set7reps, set8reps, set1weight, set2weight, set3weight, set4weight, set5weight, set6weight, set7weight, set8weight, cardio, cardio_dur, weight, comments)
VALUES ('$login_session', '$excercisename', '$setcount', '$rep1', '$rep2', '$rep3', '$rep4', '$rep5', '$rep6', '$rep7', '$rep8', '$weight1', '$weight2', '$weight3', '$weight4', '$weight5', '$weight6', '$weight7', '$weight8', '$cardio', '$duration', '$weight', '$comments')";
}
if($weight > 0){
$sql = "INSERT INTO t_workout (username, weight)
VALUES ('$login_session', '$weight')";
}
if(isset($sql)){
if ($dbcon->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Waiting for entry.";
}
}
$dbcon->close();
?>
如果設置你的體重則$ SQL重寫第一次插入。你可能想要運行兩個插入，而不是最後一個。 –
是的，我記得那一部分。我忘了確切的原因，但在網站上我指定如果你想記錄你的個人體重，留下一切空白。但即使在我的網站上輸入只有重量返回一個「等待輸入」只有我能想到的是我的桌面上的Wampserver是3.0.6，而在我的服務器上它是3.1.0。這可能是偶然的問題嗎？但是我猶豫是否認爲這只是代碼正確？所有的表格/ php都在那裏:( –
我很迷茫，我卸載了所有東西，在服務器上安裝了相同的Wampserver 3.0.6，放在桌面上的所有代碼中，以便它們應該是相同的。但是仍然會出現「正在等待輸入」的錯誤，而這個問題在桌面上並不存在。非常傷心lol。 –