未定义的数组键并且不知道该键来自哪里
undefined array key and don't know where that key comes from
我正在关注这个,
https://www.androidhire.com/insert-data-from-app-to-mysql-android/
步骤'Upload the php script on your server'有错误,
'code for get_data.php file'
<?php
include 'DatabaseConfig.php' ;
$con = mysqli_connect($HostName,$HostUser,$HostPass,$DatabaseName);
$room = $_POST['room'];
$time = $_POST['time'];
$Sql_Query = "insert into GetDataTable (room,time) values ('$room','$time')";
if(mysqli_query($con,$Sql_Query)){
echo 'Data Submit Successfully';
}
else{
echo 'Try Again';
}
mysqli_close($con);
?>
这是我的代码和我的数据库table是这个
enter image description here
错误信息是
警告:第 8 行
中 C:\xampp\htdocs\get_data.php 中未定义的数组键“room”
所以我试过了
if(isset($_POST['room']){
$room = $_POST['room'];
}
但这给了我 警告:未定义的变量,
您需要将 if()
放在使用 $room
:
的整个代码块周围
<?php
include 'DatabaseConfig.php' ;
$con = mysqli_connect($HostName,$HostUser,$HostPass,$DatabaseName);
if ( isset( $_POST['room'] ) ) {
$room = $_POST['room'];
$time = $_POST['time'];
$Sql_Query = "insert into GetDataTable (room,time) values ('$room','$time')";
if(mysqli_query($con,$Sql_Query)){
echo 'Data Submit Successfully';
} else{
echo 'Try Again';
}
mysqli_close($con);
}
?>
我正在关注这个, https://www.androidhire.com/insert-data-from-app-to-mysql-android/
步骤'Upload the php script on your server'有错误, 'code for get_data.php file'
<?php
include 'DatabaseConfig.php' ;
$con = mysqli_connect($HostName,$HostUser,$HostPass,$DatabaseName);
$room = $_POST['room'];
$time = $_POST['time'];
$Sql_Query = "insert into GetDataTable (room,time) values ('$room','$time')";
if(mysqli_query($con,$Sql_Query)){
echo 'Data Submit Successfully';
}
else{
echo 'Try Again';
}
mysqli_close($con);
?>
这是我的代码和我的数据库table是这个 enter image description here
错误信息是 警告:第 8 行
中 C:\xampp\htdocs\get_data.php 中未定义的数组键“room”所以我试过了
if(isset($_POST['room']){
$room = $_POST['room'];
}
但这给了我 警告:未定义的变量,
您需要将 if()
放在使用 $room
:
<?php
include 'DatabaseConfig.php' ;
$con = mysqli_connect($HostName,$HostUser,$HostPass,$DatabaseName);
if ( isset( $_POST['room'] ) ) {
$room = $_POST['room'];
$time = $_POST['time'];
$Sql_Query = "insert into GetDataTable (room,time) values ('$room','$time')";
if(mysqli_query($con,$Sql_Query)){
echo 'Data Submit Successfully';
} else{
echo 'Try Again';
}
mysqli_close($con);
}
?>