用PHP更新MySQL表
[英]Update MySQL table with PHP
问题描述
我有一个小脚本，可以用来更新MySQL数据库。 它首先向我显示表的所有条目，并将其放入表格中。 然后，它应将所有更改更新回数据库。 但不幸的是，该脚本无法正常工作。 谁能帮帮我吗？
这是文件：updateform.php
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"><html xmlns="http://www.w3.org/1999/xhtml"><head>
<style type="text/css">
body {
margin:50px 0px; padding:0px;
text-align:center;
font:13px Tahoma,Geneva,sans-serif
}
#content {
width:1000px;
margin:0px auto;
text-align:left;
padding:15px;
border:1px dashed #333;
background-color:#eee;
}
</style>
</head>
<body>
<div id='content'><h3><center>Update Information</center></h3>
<?php
// connect to the database
mysql_connect(host,dbuser,passwd);
// select the database
mysql_select_db(db) or die("Unable to select database");
// run the query and put the results in an array variable called $result
$result = mysql_query("SELECT * FROM program ORDER BY 'id'");
// find out how many records there are to update
$size = count($_POST['id']);
// start a counter in order to number the input fields for each record
$i = 0;
print "<table width='100%' border='0' cellspacing='1' cellpadding='0'><tr><td>";
// open a form
print "<form name='id' method='post' action='update.php'>
<table width='100%' border='0' cellspacing='1' cellpadding='1'><tr>
<td align='center'><strong>ID</strong></td>
<td align='center'><strong>Vorname</strong></td>
<td align='center'><strong>Nachname</strong></td>
<td align='center'><strong>Punkte</strong></td>
</tr>\n";
// start a loop to print all of the courses with their book information
// the mysql_fetch_array function puts each record into an array. each time it is called, it moves the array counter up until there are no more records left
while ($Update = mysql_fetch_array($result)) {
print "<tr>\n";
// assuming you have three important columns (the index (id), the course name (course), and the book info (bookinfo))
// start displaying the info; the most important part is to make the name an array (notice bookinfo[$i])
print "<td align='center'><p>{$Update['id']}</p></td>\n";
print "<td align='center'><input type='text' size='40' name='vorname[$i]' value='{$Update['vorname']}' /></td>\n";
print "<td align='center'><input type='text' size='40' name='nachname[$i]' value='{$Update['nachname']}' /></td>\n";
print "<td align='center'><input type='text' size='40' name='punkte[$i]' value='{$Update['punkte']}' /></td>\n";
print "</tr>\n";
// add 1 to the count, close the loop, close the form, and the mysql connection
++$i;
}
print "<tr>
<td colspan='4' align='center'><input type='submit' value='submit' />";
print "</td>
</tr>
</table>
</td>
</tr>
</form>
</table>";
mysql_close();
?><br /><br /><div></body>
</html>
这是文件：update.php
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"><html xmlns="http://www.w3.org/1999/xhtml"><head>
<style type="text/css">
body {
margin:50px 0px; padding:0px;
text-align:center;
font:13px Tahoma,Geneva,sans-serif
}
#content {
width:1000px;
margin:0px auto;
text-align:left;
padding:15px;
border:1px dashed #333;
background-color:#eee;
}
</style>
</head>
<body>
<div id='content'><h3><center>Success! </center></h3>
<table width='100%' border='0' cellspacing='1' cellpadding='0'><tr><td>
<table width='100%' border='0' cellspacing='1' cellpadding='1'>
<tr>
<td align='center'><strong>Vorname</strong></td>
<td align='center'><strong>Nachname</strong></td>
<td align='center'><strong>Punkte</strong></td>
<td align='center'><strong>ID</strong></td>
</tr>
<?php
// connect to the database and select the correct database
mysql_connect(host,dbuser,passwd);
mysql_select_db(db) or die("Unable to select database");
// find out how many records there are to update
$size = count($_POST['id']);
// start a loop in order to update each record
$i = 0;
while ($i < $size) {
// define each variable
$vorname = $_POST['vorname'][$i];
$nachname = $_POST['nachname'][$i];
$punkte = $_POST['punkte'][$i];
$id = $_POST['id'][$i];
// do the update and print out some info just to provide some visual feedback
$query = "UPDATE `program` SET `vorname` = '$vorname', `nachname` = '$nachname', `punkte` = '$punkte' WHERE `id` = '$id' LIMIT 1";
mysql_query($query) or die ("Error in query: $query");
print "
</tr>
<td align='left'><p>$vorname</p></td>
<td align='left'>$nachname</td>
<td align='left'>$punkte</td>
<td align='left'>$id</td>
</tr>
";
++$i;
}
mysql_close();
?>
<tr>
<td colspan='4' align='center'>
</td>
</tr>
</table>
</td>
</tr>
</table></div></body></html>
有人可以帮我吗？ 问候亚历克斯
2 个解决方案
解决方案1
1 已采纳 2013-05-30 22:58:22
更改
print "<td align='center'><p>{$Update['id']}</p></td>\n";
至
print "<td align='center'>
<p>{$Update['id']}</p>
<input type=\"hidden\" name=\"id[$i]\" value=\"{$Update['id']}\" />
</td>\n";
解决方案2
0 2013-05-30 23:11:50
问题是您没有正确跟踪记录的ID。
尝试根据ID命名表单文本输入。
例如 ：
print "<td align='center'><input type='text' size='40' name='vorname_".$Update['id']."' value='".$Update['vorname']."' /></td>\n";
之后，您可以使用explode，strval或regex函数删除id。
PHP，MySQL更新表
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