我需要将记录添加到表“writings”中。正如您在下面的代码中看到的,失败导致了消息 data insert failed
。
我已经尝试更改在线论坛建议的所有内容,但没有帮助。所有程序输出都是 data insert failed
,仅此而已。当我尝试添加 mysql_errno
时,它给出了错误代码 0
,我记得这意味着操作成功。
主要代码:
<?php
include 'dbcon.php';
$Title=$_GET['Title'];
$TaskID=$_GET['TaskID'];
$uid=$_GET['uid'];
$Text=$_GET['Text'];
$Date=$_GET['Date'];
$result=mysqli_query($MySQLiconn,"insert into 'writings'(uid,TaskID,Date,Text,Title, comment) values('$uid','$TaskID','$Date','$Text','$Title','')");
if ($result) {
echo "Successfully added.<br>";
} else {
echo "Data insert failed <br><br>";
}
文件DBcon
<?php
$DB_host = "localhost";
$DB_user = "root";
$DB_pass = "";
$DB_name = "web";
$MySQLiconn = new MySQLi($DB_host,$DB_user,$DB_pass,$DB_name);
if($MySQLiconn->connect_errno)
{
die("ERROR : -> ".$MySQLiconn->connect_error);
}
最佳答案
试试这个:
$sql = "INSERT INTO writings (uid, TaskID, Date, Text, Title, comment)
VALUES ('$uid','$TaskID','$Date','$Text','$Title','')";
$result = mysql_query( $sql, $MySQLiconn );
if(! $result )
{
die('Could not enter data: ' . mysql_error());
}
echo "Entered data successfully\n";
关于php - 在 php mysql 中插入数据失败,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/43773309/