用PHP更新多行
[英]Updating multiple rows with PHP
问题描述
I want to export my mysql-database into a php-form, and update it, if I make any changes to the input-boxes, through the submit-button. 
我想将我的mysql-database导出到一个php-form中，并通过“提交”按钮对输入框进行任何更改，以对其进行更新。 My main-problem is, that I want to update multiple rows at once. 我的主要问题是，我想一次更新多行。 Here is my code; 这是我的代码； I don't get any errors, it just doesn't work. 我没有任何错误，只是没有用。
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Frameset//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-frameset.dtd">
<html>
<head>
<style>
<!--
A:link {text-decoration: none; color:#005BA2}
A:visited {text-decoration: none; color:#005BA2}
A:active {text-decoration: underline; color:#FF0000}
A:hover { color: #FF0000; line-height: normal; text-decoration: underline; background-color: #FFFFFF }
input.center { text-align:center; }
body { color: #0061AE; font-size: 13px; font-family: Arial, Helvetica, sans-serif; }
-->
</style>
</head>
<body>
<center>
<form method="post" action="#">
<?php
// Fehlermeldungen ausschalten wegen undef. Variablen
// error_reporting(E_ERROR | E_PARSE);
// Datenbankvariablen setzen
$hostname = "localhost";
$database = "intranet";
$username = "intranet";
$password = "intranet";
$table = "arbeitsanweisungen";
// Verbindung mit Datenbank herstellen
$db_connect = mysql_connect($hostname, $username, $password);
mysql_select_db("$database") or die();
// Datenbankabfrage
$sql = "SELECT * FROM $table ORDER BY kennung";
$result = mysql_query($sql);
// Zeilen zählen
$count = mysql_num_rows($result);
?>
<table border="1" cellspacing="0" cellpadding="2">
<tr>
<td align="center"><strong>ID</strong></td>
<td align="center"><strong>Kennung</strong></td>
<td align="center"><strong><img src="images/german.jpg"></strong></td>
<td align="center"><strong><img src="images/usa.jpg"></strong></td>
<td align="center"><strong>deutscher Titel</strong></td>
<td align="center"><strong>englischer Titel</strong></td>
<td align="center"><strong>Mitarbeiter</strong></td>
</tr>
<?php
while($rows=mysql_fetch_array($result)){
?>
<tr>
<td align="center">
<?php
$id[]=$rows['id']; ?><?php echo $rows['id'];
?>
</td>
<td align="center">
<input name="kennung[]" size="15" type="text" id="kennung" value="<?php echo $rows['kennung']; ?>">
</td>
<td align="center">
<input name="german[]" class="center" size="1" type="text" id="german" value="<?php echo $rows['german']; ?>">
</td>
<td align="center">
<input name="english[]" class="center" size="1" type="text" id="english" value="<?php echo $rows['english']; ?>">
</td>
<td align="center">
<input name="nameDE[]" size="50" type="text" id="nameDE" value="<?php echo $rows['nameDE']; ?>">
</td>
<td align="center">
<input name="nameEN[]" size="50" type="text" id="nameEN" value="<?php echo $rows['nameEN']; ?>">
</td>
<td align="center">
<input name="mitarbeiter[]" size="25" type="text" id="mitarbeiter" value="<?php echo $rows['mitarbeiter']; ?>">
</td>
</tr>
<br>
<?php
};
?>
</table>
<input type="submit" name="Submit" value="Submit">
<?php
if(isset($_POST['Submit'])){
$kennung = $_POST['kennung'];
$german = $_POST['german'];
$english = $_POST['english'];
$nameDE = $_POST['nameDE'];
$nameEN = $_POST['nameEN'];
$mitarbeiter = $_POST['mitarbeiter'];
for($i=0;$i<$count;$i++){
mysql_query("UPDATE $table SET kennung = '$kennung[$i]', german = '$german[$i]', english = '$english[$i]', nameDE = '$nameDE[$i]', nameEN = '$nameEN[$i]', mitarbeiter = '$mitarbeiter[$i]' WHERE id = '$id[$i]'");
}
}
// Verbindung schließen
mysql_close();
?>
</form>
</center>
</body>
</html>
4 个解决方案
解决方案1
1 2014-03-18 10:09:58
In side if(isset($_POST['Submit'])){ } condition, please write following code and check again: 在if（isset（$ set _ $ [POST_'Submit']））{}条件下，请编写以下代码并再次检查：
for($i=0;$i<$count;$i++){
mysql_query("UPDATE $table SET kennung = '".$_POST[$kennung[$i]]."', german = '".$_POST[$german[$i]]."', english = '".$_POST[$english[$i]]."', nameDE = '".$_POST[$nameDE[$i]]."', nameEN = '".$_POST[$nameEN[$i]]."', mitarbeiter = '".$_POST[$mitarbeiter[$i]]."' WHERE id = '".$id[$i]."'");
}
解决方案2
0 2014-03-18 09:36:46
Don't put arrays directly into a quoted string: 不要将数组直接放在带引号的字符串中：
mysql_query("UPDATE $table SET kennung = '$kennung[$i]', ger....")
Properly concatenate it using a . 使用正确连接它. : ：
mysql_query("UPDATE " . $table . " SET kennung = '" . $kennung[$i] . "', ger..");
Also, enable error reporting: 另外，启用错误报告：
<?php
// Put these lines to the top of your script
error_reporting(E_ALL);
ini_set('display_errors', true);
ini_set('display_startup_errors', true);
ini_set('xmlrpc_errors', true);
And check for mysql_last_error(); 并检查mysql_last_error(); or mysql_error(); 或mysql_error(); (not sure atm). （不确定atm）。
mysql_* is deprecated, use mysqli or PDO . mysql_*已过时，请使用mysqli或PDO 。
解决方案3
0 2014-03-18 09:41:30
You need to put $rows['id'] into the form: 您需要将$rows['id']放入以下形式：
<td align="center">
<?php echo $rows['id'] . '<input type="hidden" name="id[]" value="' . $rows['id'] . '">';
?>
</td>
Then when you're processing the form, you have to use 然后，当您处理表格时，您必须使用
$id = $_POST['id'];
解决方案4
0 2014-03-18 09:53:03
if you want to update multiple rows in a single query than you need to use WHERE id IN (); 如果要multiple rows in a single query更新multiple rows in a single query需要使用WHERE id IN (); in your query ... 在您的query ...
so your query will be like 所以你的查询就像
UPDATE $table SET kennung = '$kennung[$i]', german = '$german[$i]', english = '$english[$i]', nameDE = '$nameDE[$i]', nameEN = '$nameEN[$i]', mitarbeiter = '$mitarbeiter[$i]' WHERE id IN ($id_array)
and your $id_array will contain all the ids which needs to be updated 并且您的$id_array将包含所有ids which needs to be updated的ids which needs to be updated
like 喜欢
$id_array = array(`1,2,3,4,5`);
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