Question

我想用url发送数据我的代码php是：

<?php    
$db_host="127.0.0.1"; $db_uid="root"; $db_pass="";     
$db_name="highway_db";    $db_con =    
mysqli_connect($db_host,$db_uid,$db_pass,$db_name);
$longitude =(isset( $_GET['lon`enter code here`gitude']));
$latitude = (isset($_GET['latitude']));
$timestamp = (isset($_GET['timestamp']));
$result = mysqli_query($db_con,"INSERT INTO accident (longitude, latitude,   timestamp) 
VALUES ($longitude, $latitude, $timestamp)");
if($result == true)
echo '{"query_result":"SUCCESS"}';
else
echo '{"query_result":"FAILURE"}';
mysqli_close($db_con);
?>

我的表是偶然的，有经度，经度，纬度和时间戳 id是AUTO_INCREMENT。但是当我使用这个网址时出现问题：

http://127.0.0.1/addAccidents.php?longitude=3.54&latitude=3.09&timestamp=2016-04-25 11:11:00 
i find that is add to my table accident 
longitude=1
latitude =1,
timestamp =0000-00-00 00:00:00.

这是我的问题，请帮助我

Answer 1

试试这个：

 $result = mysqli_query($db_con,"INSERT INTO accident (longitude, latitude, timestamp) VALUES ('".$longitude."', '".$latitude."', '".$timestamp."')");

回显sql查询并尝试直接在phpmyadmin中运行查询，并检查数据库表列类型。

Answer 2

尝试改变hte变量的设置 - 无论如何你应该这样做，因为即使GET值不存在 - 这段代码仍然会尝试将值插入到db中，例如：

if(isset( $_GET['longitude'])){$longitude =$_GET['longitude']};

然后编写代码，只有在设置了三个值时才允许写入db。

Answer 3

isset() is a function which checks if a given variable has contents or not. If it has it returns 1 and if not it return 0.  

In all 3 below statements, you are just checking the variables, you have not assigned  values to your variables!

Try to use this code here, it should work if GET REQUEST with those specified parameters is sent!

<?php    
$db_host="127.0.0.1"; 
$db_uid="root"; 
$db_pass="";     
$db_name="highway_db";    
$db_con = mysqli_connect($db_host,$db_uid,$db_pass,$db_name);


$longitude = isset($_GET['longitude'])?$_GET['longitude']:"";
$latitude = isset($_GET['latitude'])?$_GET['latitude']:"";
$timestamp = isset($_GET['timestamp'])?$_GET['timestamp']:"";

$query = "INSERT INTO accident (longitude,latitude,timestamp) VALUES ($longitude,$latitude,$timestamp)";

if(mysqli_query($db_con,$query)){
echo '{"query_result":"SUCCESS"}';
}
else{
echo '{"query_result":"FAILURE"}';
echo "ERROR= ". mysql_error();
}
mysqli_close($db_con);
?>
Let me know how it goes....

url没有发送正确的数据，我可以做什么来发送正确的值

3 个答案: