我一直在爲一個論壇寫一些代碼,並且是PHP新手,但我遇到了一些麻煩。 當我通過輸入答案測試程序時,我收到一個顯示「ERROR」的網頁。論壇響應錯誤:重複項PRIMARY鍵
改變echo "ERROR"
到echo mysql_error()
後,網頁改成這樣:
Notice: Undefined index: id in C:\xampp5\htdocs\add_answer.php on line 14
Notice: Undefined index: a_name in C:\xampp5\htdocs\add_answer.php on line 30
Notice: Undefined index: a_email in C:\xampp5\htdocs\add_answer.php on line 31
Notice: Undefined index: a_answer in C:\xampp5\htdocs\add_answer.php on line 32 Duplicate entry '1' for key 'PRIMARY'
<?php
$host="localhost"; // Host name
$username=""; // Mysql username
$password=""; // Mysql password
$db_name="test"; // Database name
$tbl_name="forum_answer"; // Table name
// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
// Get value of id that sent from hidden field
$id = $_POST['id'];
// Find highest answer number.
$sql="SELECT MAX(a_id) AS Maxa_id FROM $tbl_name WHERE question_id='$id'";
$result=mysql_query($sql);
$rows=mysql_fetch_array($result);
// add + 1 to highest answer number and keep it in variable name "$Max_id". if there no answer yet set it = 1
if ($rows) {
$Max_id = $rows['Maxa_id']+1;
}
else {
$Max_id = 1;
}
// get values that sent from form
$a_name = $_POST['a_name'];
$a_email = $_POST['a_email'];
$a_answer = $_POST['a_answer'];
$datetime=date("d/m/y H:i:s"); // create date and time
// Insert answer
$sql2="INSERT INTO $tbl_name(question_id, a_id, a_name, a_email, a_answer, a_datetime)VALUES('$id', '$Max_id', '$a_name', '$a_email', '$a_answer', '$datetime')";
$result2=mysql_query($sql2);
if($result2){
echo "Successful<BR>";
echo "<a href='view_topic.php?id=".$id."'>View your answer</a>";
// If added new answer, add value +1 in reply column
$tbl_name2="forum_question";
$sql3="UPDATE $tbl_name2 SET reply='$Max_id' WHERE id='$id'";
$result3=mysql_query($sql3);
}
else {
echo "ERROR";
}
// Close connection
mysql_close
();
?>
對於我不關心安全的那一刻,我打算以後在軌道下解決安全問題。
**警告**:如果您只是學習PHP,請不要使用['mysql_query'](http://php.net/manual/en/function.mysql-query.php)接口。這是非常可怕和危險的,它在PHP 7中被刪除了。[PDO的替代品並不難學](http://net.tutsplus.com/tutorials/php/why-you-should-be-using-phps -pdo-for-database-access /)以及[PHP The Right Way](http://www.phptherightway.com/)等指南介紹了最佳實踐。你的用戶數據是**不是** [正確轉義](http://bobby-tables.com/php.html),並有[SQL注入漏洞](http://bobby-tables.com/),並且可以被利用。 – tadman
打印出正確的錯誤信息而不是無用的錯誤。 – Shadow
用'echo mysql_error()'替換'echo「ERROR」'並將結果編輯到您的問題中。 –