Question

可能重复：
PHP: “Notice: Undefined variable” and “Notice: Undefined index”

我知道它的基本但是我已经进行了8个小时的编码，这让我陷入困境。为我的更新查询接收第14行上所有帖子的未定义索引。感谢您对此事的任何帮助。

<html>
<head>  
</head>
<body>
<h1><center><i>Edit/Add Course ID</i></center></h1>
<?php 
$con = mysql_connect("localhost","u/n","p/w");
if (!$con){
die ("Can not connect: " . mysql_error());
}
mysql_select_db("my_db",$con);

if(isset($_POST['submit'])){
$UpdateQuery = "UPDATE course SET cours_num='$_POST[coursnum]', 
cours_title='$_POST[courstitle]', cours_desc='$_POST[coursdesc]' WHERE 
cours_num='$_POST[hidden]'";
mysql_query($UpdateQuery, $con);

};
$sql = "Select * FROM course";
$myData = mysql_query($sql,$con);

while($record = mysql_fetch_array($myData)){
echo "<form action=mydata.php method=post>";
echo "<b>Course Number:</b></br>";
echo "<input type=text maxlength=10 size=10 readonly name=coursnum value=" . 
$record['cours_num'].">";
echo "</br></br>";
echo "<b>Course Title:</b><br/>";
echo "<input type=text maxlength=10 size=10 name=courstitle value=" . 
$record['cours_title'].">";
echo "</br></br>";
echo "<b>Course Description:</b></br>";
echo "<input type=text name=coursdesc max length=50 size=50 value=" . 
$record['cours_desc'].">";
echo "</br></br>";
echo "<input type=hidden name=hidden value=" . $record['cours_num'].">";
echo "<center><input type=submit name=submit value=Edit></center>";
echo "</form>";
}

mysql_close($con);

?>
</body>
</html>

Answer 1

由于以下行

，您收到错误

$UpdateQuery = "UPDATE course SET cours_num='$_POST[coursnum]', 
cours_title='$_POST[courstitle]', cours_desc='$_POST[coursdesc]' WHERE 
cours_num='$_POST[hidden]'";

您可以使用

$UpdateQuery = sprintf("UPDATE course SET cours_num='%d' , 
                cours_title='%s', 
                cours_desc='%s',
                WHERE  cours_num = '%d'", mysql_real_escape_string($_POST['coursnum']), mysql_real_escape_string($_POST['courstitle']), mysql_real_escape_string($_POST['coursdesc']), mysql_real_escape_string($_POST['hidden']));

您还应确保始终设置变量

示例

$_POST['coursnum'] = isset($_POST['coursnum']) ? $_POST['coursnum'] : null ;

php代码第14行中未定义的索引

1 个答案: