在php中未定义的索引错误，虽然我定义它。为什么？

我新的PHP所以请指导，不让它重复，因为我无法摆脱以前的solutions.My PHP代码解决方案如下在php中未定义的索引错误，虽然我定义它。为什么？

<?php 
if($_SERVER['REQUEST_METHOD']=='GET'){ 
$city = $_GET['city']; 
$town = $_GET['town']; 
//$skills=$_POST['skills']; 
require_once('DbConnect.php'); 
//Creating sql query 
$sql = "SELECT FROM employees where city='".$city."' and town='".$town."'"; 

//getting result 
$r = mysqli_query($con,$sql); 

//creating a blank array 
$result = array(); 

//looping through all the records fetched 
while($row = mysqli_fetch_array($r)){ 

//Pushing name and id in the blank array created 
array_push($result,array(
"name"=>$row['name'], 
"phone"=>$row['phone'], 
"skills"=>$row['skills'] 
)); 
} 

//Displaying the array in json format 
echo json_encode(array('result'=>$result)); 

mysqli_close($con); 
} 
?>

错误

给出注意：未定义指数：城市在C：\ XAMPP \ htdocs中\ getServices \ employeesInfo.php上线3

注意：未定义指数：镇在C：\ XAMPP \ htdocs中\ getServices \ employeesInfo.php 4号线

警告：mysqli_fetch_array（）预计参数1被mysqli_result，在C中给出布尔：\ XAMPP \ htdocs中\ getServices \ employeesInfo.php上线17 { “结果”：[]}

来源

2017-09-10 A.Mateen

将您的SQL查询更改为'$ sql =“SELECT * FROM employees ...' –

什么是URL或表单？你开放SQL注入，参数化。只检查请求方法是不够的。 – chris85

除了显而易见的SQL注入问题，如果没有在您的URL中定义城市或城镇的变量，那么您将得到此错误 –

需要通过做选择一个或多个列，例如选择所有SELECT * FROM..，查询应该是这样的

$sql = "SELECT * FROM employees where city='".$city."' and town='".$town."'";

Update_Code：

<?php 

if($_SERVER['REQUEST_METHOD']=='GET' && !empty($_GET['city']) && !empty($_GET['town'])){ 

$city = $_GET['city']; 
$town = $_GET['town']; 

//$skills=$_POST['skills']; 
require_once('DbConnect.php'); 

//Creating sql query 
$sql = "SELECT * FROM employees where city='".$city."' and town='".$town."'"; 

//getting result 
$r = mysqli_query($con,$sql); 

//creating a blank array 
$result = array(); 

//looping through all the records fetched 
while($row = mysqli_fetch_array($r)){ 

//Pushing name and id in the blank array created 
array_push($result,array(
"name"=>$row['name'], 
"phone"=>$row['phone'], 
"skills"=>$row['skills'] 
)); 
} 

//Displaying the array in json format 
echo json_encode(array('result'=>$result)); 

mysqli_close($con); 
} 
else{ 

$message = " Fill all the details First"; 

} 

if(isset($message) && !empty($message)){ 

echo "$message"; 
} 

?>

来源

2017-09-10 17:14:21

在php中未定义的索引错误，虽然我定义它。为什么？

回答

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