php 注意：未定义索引：id in C:\xampp\htdocs\
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Notice: Undefined index: id in C:\xampp\htdocs\
提问by Wong Jia Min Jacey
Possible Duplicate:
Reference - What does this error mean in PHP?
可能的重复：
参考 - 这个错误在 PHP 中意味着什么？
I am trying to establish a database using PHP but I keep getting errors saying I have an undefined index in id.
我正在尝试使用 PHP 建立一个数据库，但我不断收到错误消息，说我在 id 中有一个未定义的索引。
I want to add data into my database, also able to update, delete and show the infomation. This is my update info codes.
我想将数据添加到我的数据库中，还可以更新、删除和显示信息。这是我的更新信息代码。
<?PHP
// Connection to MySQL
$dbconnection = @mysql_connect('localhost','root','');
if (!$dbconnection) {
echo '<p> Unable to connect to the database at this time.</br></br></p>';
exit();}
else {
echo '<p> connection to database is successful</br></br> </p>';}
//select Mysql Database-ijdb
if (!@mysql_select_db('disease')){
exit('<p> Unable to locate the information on the database.</p>');
}
//Receive Variables from the GET of JOKELIST.php
if(isset($_POST['submit']))
{
$GeneticOrganisation=$_POST['newGeneticOrganisation'];
$ProteinInformation=$_POST['newProteinInformation'];
$Symptoms=$_POST['newSymptoms'];
$Population=$_POST['newPopulation'];
$Cure=$_POST['newCure'];
$OriginOfDisease=$_POST['newOriginOfDisease'];
$dmdid=$_POST['id'];
// Print receieved variables to check accuracy
$sql= "UPDATE dmd SET GeneticOrganisation ='".$GeneticOrganisation."' WHERE id
`='".$dmdid."'";
$sql= "UPDATE dmd SET ProteinInformation ='".$ProteinInformation."' WHERE id ='".$dmdid."'";
$sql= "UPDATE dmd SET Symptoms ='".$Symptoms."' WHERE id ='".$dmdid."'";
$sql= "UPDATE dmd SET Population ='".$Population."' WHERE id ='".$dmdid."'";
$sql= "UPDATE dmd SET Cure ='".$Cure."' WHERE id ='".$dmdid."'";
$sql= "UPDATE dmd SET OriginOfDisease ='".$OriginOfDisease."' WHERE id ='".$dmdid."'";
}
if (!@mysql_query($sql))
echo "<p> Information could not be updated-".mysql_error();
else{
echo "<p> Information updated successfully";
echo '<a href="diseaseInfo.php"> View the information on the disease here.';
}
?>
回答by EM-Creations
It sounds like $_POST['id']isn't being set.
好像$_POST['id']没有设置
You need to make sure it's being set before using it, or check if it's set and give it a default value if it isn't.
在使用之前，您需要确保它已设置，或者检查它是否已设置，如果未设置，则为其提供默认值。
Eg:
例如：
if (isset($_POST['id'])) { // If the id post variable is set
$dmid = $_POST['id'];
} else { // If the id post variable is not set
$dmid = 1;
}
回答by mvbrakel
The following line will most likelt cause it:
以下行最有可能导致它：
$dmdid=$_POST['id'];
Verify your form (or other POST supplier) provides the 'id' key. with something like:
验证您的表单（或其他 POST 供应商）提供了“id”键。像这样：
<input type="hidden" name="id" value="xx"/>
