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任何想法为什么下面的代码在用户填写表单后没有向数据库添加任何内容?我真的很感激它。PHP&MySQL INSERT INTO问题
谢谢!
if($_SESSION['loginSuccess']==1) {
// ============================================================
// = Create the table of current tasks stored in the database =
// ============================================================
$userID = $_SESSION['userID'];
$result = mysql_query("SELECT * FROM Tasks WHERE userID = '$userID'");
echo "<div id=\"draggable\" class=\"ui-widget-content\"><table border='5'><tr class=\"ui-widget-header\"><td><u>Task Name:</u></td><td><u>Class:</u></td><td><u>Due Date:</u></td><td><u>Task Type:</u></td></tr>";
echo $_SESSION['userID'];
while($row = mysql_fetch_array($result)) {
$taskName = $row[1];
$class = $row[2];
$taskDueDate = $row[3];
$taskType = $row[4];
echo "<tr><td>'$taskName'</td><td>'$class'</td><td>'$taskDueDate'</td><td>'$taskType'</td></tr>";
}
echo "</table>";
function addNewTask ($name, $class, $dueDate, $type) {
mysql_query("INSERT INTO Tasks VALUES ('$userID','$name', '$class', '$dueDate', '$type')");
}
if($_POST['taskName'] != NULL) {
addNewTask($_POST['taskName'], $_POST['class'], $_POST['dueDate'], $_POST['dueDate']);
}
?>
<!-- <img border="1" alt="New" src="/newTask.png" id="newTask" onmouseClick="showTaskField"/> -->
<p><form name="newTask" method="post" action="index.php" id="newTask"><br>
Task Name: <input name="taskName" type="text"> (necessary)<br>
Class: <input name="class" type="text"><Br>
Due Date: <input name="dueDate" type="text" id="datepicker"><Br>
Type:
<input type="submit"></p></div>
哇,哇,哇! [SQL注入](http://www.tizag.com/mysqlTutorial/mysql-php-sql-injection.php)警报!请仔细阅读以确保您的查询(我认为使用[PDO](http://php.net/pdo)和[预准备语句](http://php.net/manual/en/pdo.prepared-statements.php )是最好的解决方案,但有一些),然后检查1)是否正在运行addNewTask? 2)查询是否返回错误? (打印`mysql_error()`找出) – Matchu 2010-12-02 03:03:31
调试你的mysql_query`调用,通过检查它是否返回true或不使用`mysql_error`来判断返回false的错误。 – deceze 2010-12-02 03:03:51