我是php的新手,所以请指导,不要让它重复,因为我无法从以前的解决方案中获得解决方案。我的PHP代码如下所示
<?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
注意:未定义的索引:第3行的C:\ xampp \ htdocs \ getServices \ employeesInfo.php中的城市
注意:未定义的索引:第4行的C:\ xampp \ htdocs \ getServices \ employeesInfo.php中的town
警告:mysqli_fetch_array()要求参数1为mysqli_result,第17行的C:\ xampp \ htdocs \ getServices \ employeesInfo.php中给出布尔值 { “结果”:[]}
答案 0 :(得分:1)
您需要选择一个或多个列,例如通过全选 SELECT * FROM..
,查询将如下所示
$sql = "SELECT * FROM employees where city='".$city."' and town='".$town."'";
<强> Update_Code:强>
<?php
if($_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";
}
?>