这是我的代码:
<?php
$host="localhost"; // Host name
$username="root"; // Mysql username
$password="blah"; // Mysql password
$db_name="test"; // Database name
$tbl_name="SubCategories"; // Table name
$con=mysqli_connect("$host", "$username", "$password", "$db_name");
if (mysqli_connect_errno()) // Check connection
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
?>
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8" />
<title></title>
</head>
<body>
<form action="untitled.php" method="post"><!-- untitled.php -->
<?php
//print_r($_POST); //print all checked elements
//echo "<br>".$email, $_POST["update"][$i];
//mysql_real_escape_string ($route )
if(isset($_POST['submit'])) {
foreach ($_POST["holder"] as $i=>$email) {
$y=$email;
$h=$_POST["update"][$i];
$res2=mysqli_query("UPDATE ".$tbl_name." SET subCat2 = '" . $y . "' WHERE id =". $h,$con);
if ($res2){
}
else{
echo "<h1>NOT WORKING!</h1>";
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
}
}
$result = mysqli_query($con,"SELECT * FROM $tbl_name");
echo "<br>";
while($row = mysqli_fetch_array($result))
{
echo '<input type="text" name="holder[]" id="checkbox-1" class="custom" value=" ' . $row['subCat2'] . '"/>';
echo '<input type="hidden" name="update[]" id="checkbox-1" class="custom" value=" ' . $row['subCatNum'] . '"/>';
echo "<br>";
}
?>
</br>
<input type="submit" name="submit">
</form>
</body>
</html>
我无法更新数据库中的表。我能够正确提取变量并回显它们,但是它不起作用。
我在过去“未选择数据库”中遇到了以下错误。
最佳答案
我认为您忘记选择数据库。尝试将其放在连接之后:
if (!mysqli_select_db($con, $db_name)) {
die("Uh oh, couldn't select database $db_name");
}
如果发生这种情况,请仔细检查名称、权限等。
关于php - 为什么不更新到数据库?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/20279215/