我尝试更新表showtable
中的一行Bugupdate 通过使用下面的php代码,将bugID绑定到SQL UPDATE语句以更新我想要的行,但它似乎不起作用,问题在于我的SQL语句吗?
$id = $_GET['update'];
$games = htmlentities($_POST['games']);
$version = htmlentities($_POST['version']);
$platform = htmlentities($_POST['platform']);
$frequency = htmlentities($_POST['frequency']);
$proposal = htmlentities($_POST['proposal']);
$SQLstring2 = "UPDATE " .$TableName. " SET Game=?,Version=?,Platform=?,Frequency=?,Proposed solution=? WHERE BugID= " .$id;
if ($stmt = mysqli_prepare($DBconnect, $SQLstring2)) {
mysqli_stmt_bind_param($stmt,'sssss', $games, $version, $platform, $frequency, $proposal);
$QueryResult2 = mysqli_stmt_execute($stmt);
if ($QueryResult2 === FALSE) {
echo "<p>Unable to execute the query.</p>"
. "<p>Error code "
. mysqli_errno($DBconnect)
. ": "
. mysqli_error($DBconnect)
. "</p>";
} else {
echo "<h1> Thank you for your contribution";
}
mysqli_stmt_close($stmt);
}
mysqli_close($DBconnect);
答案 0 :(得分:-1)
尝试将Proposed solution
列重命名为Proposed_solution
并调整sql查询,如下所示:
$SQLstring2 = "UPDATE " .$TableName. " SET Game=?,Version=?, Platform=?, Frequency=?, Proposed_solution=? WHERE BugID= " .$id;