[英]Form not updating database
我有一個從員工列表中填充的更新表單。 值正在傳遞但未在數據庫中更新。 這是我的代碼以及我所顯示的內容。
<?php
$con = mysql_connect("localhost","root","*******");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
$query = mysql_query("select * from backup");
if(isset($_POST['update']))
$id = $_POST['id'];
$first = $_POST['first'];
$last = $_POST['last'];
$store = $_POST['store'];
$title = $_POST['title'];
$title2 = $_POST['other'];
$phone = $_POST['phone'];
$email = $_POST['email'];
$dept = $_POST['dept'];
$bio = $_POST['bio'];
$query="UPDATE backup SET first='$first', last='$last', store='$store', title='$title', title2='$title2', phone='$phone', email='$email', bio='$bio' WHERE id='$id'";
mysql_query($query);
echo "Record Updated";
mysql_close();
print_r($_POST)
?>
這是結果
記錄UpdatedArray([id] => 1396 [first] => Charles [last] => Adams [store] => [dept] =>會計[title] =>會計文員[其他] => [電話] => 410 -555-1212 [email] => [email2] => [bio] =>查理於2009年8月開始。這是一個測試...... [提交] =>提交)
有人可以幫我解決我可能做錯的事嗎? 至於注射,我將在完成測試后修復它。 我知道這聽起來可能倒退,但我需要找出為什么這不起作用。
感謝您的幫助
您還應該選擇數據庫:
// make foo the current db
$db_selected = mysql_select_db('foo', $link);
if (!$db_selected) {
die ('Can\'t use foo : ' . mysql_error());
}
但最重要的是,你應該使用MySQLi或PDO_MySQL。
這是因為,你忘了{
after if (isset(...))
。
此外,未選擇任何數據庫。
更正后的代碼如下:
<?php
$con = mysql_connect("localhost","root","*******");
mysql_select_db('DB_NAME');
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
$query = mysql_query("select * from backup");
if(isset($_POST['update'])) {
$id = $_POST['id'];
$first = $_POST['first'];
$last = $_POST['last'];
$store = $_POST['store'];
$title = $_POST['title'];
$title2 = $_POST['other'];
$phone = $_POST['phone'];
$email = $_POST['email'];
$dept = $_POST['dept'];
$bio = $_POST['bio'];
$query="UPDATE backup SET first='$first', last='$last', store='$store', title='$title', title2='$title2', phone='$phone', email='$email', bio='$bio' WHERE id='$id'";
mysql_query($query);
echo "Record Updated";
mysql_close();
print_r($_POST)
}
?>
您尚未在連接中定義數據庫。
您的代碼可以更正為:
<?php
$con = mysqli_connect("localhost","root","**","db");
if (!$con)
{
die('Could not connect: ' . mysqli_error());
}
//$query = mysqli_query($con,"select * from backup");
if(isset($_POST['update'])):
$id = $_POST['id'];
$first = $_POST['first'];
//other stuffs
$query="UPDATE backup SET first='$first', last='$last', store='$store', title='$title', title2='$title2', phone='$phone', email='$email', bio='$bio' WHERE id='$id'";
$res = mysqli_query($con,$query);
if(!$res)
die("could not update records. Error = ".mysqli_error($con));
echo "Record Updated";
mysqli_close($con);
print_r($_POST);
endif;
?>
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