[英]PHP Mysql update query returns nothing
Hi guys my php update query doesnt return me any value. 嗨,大家好,我的php更新查询没有返回任何值。 It should return me success
or failed
but its not can you guys fix this? 它应该使我success
还是failed
但是你们不能解决这个问题吗?
disregard securities here I just use this query for my android app. 在这里不考虑证券,我只对我的Android应用程序使用此查询。
Here is my code. 这是我的代码。
<?php
include_once("connection.php");
if(isset($_POST['txtCar_No']) && isset($_POST['txtCarModel']) &&
isset($_POST['txtCarType']) && isset($_POST['txtCapacity']) &&
isset($_POST['image']) && isset($_POST['txtFuelType']) &&
isset($_POST['txtPlateNumber']) && isset($_POST['txtcarPrice']))
{
$now = DateTime::createFromFormat('U.u', microtime(true));
$id = $now->format('YmdHis');
$upload_folder = "upload";
$path = "$upload_folder/$id.jpeg";
$fullpath = "http://carkila.esy.es/$path";
$image = $_POST['image'];
$Car_No = $_POST['txtCar_No'];
$Car_Model = $_POST['txtCarModel'];
$Car_Type = $_POST['txtCarType'];
$Capacity = $_POST['txtCapacity'];
$Fuel_Type = $_POST['txtFuelType'];
$PlateNumber = $_POST['txtPlateNumber'];
$carPrice = $_POST['carPrice'];
$query = "UPDATE tbl_cars SET Car_Model='$Car_Model', Car_Type='$Car_Type', Capacity='$Capacity', fuelType='$Fuel_Type' ,carPlatenuNumber='$PlateNumber', image='$fullpath' , carPrice = '$carPrice' WHERE Car_No=$Car_No";
$result = mysqli_query($conn,$query);
echo $Car_No;
if($result > 0){
echo "success";
exit();
} else {
echo "failed";
exit();
}
}
?>
You have to use mysqli_affected_rows($conn)
to get rows affected by this update query. 您必须使用mysqli_affected_rows($conn)
来获取受此更新查询影响的行。
Eg: 例如:
$result = mysqli_query($conn,$query);
$count = mysqli_affected_rows($conn);
if($result == TRUE && $count > 0){
echo "success";
exit();
} else {
print_r (mysqli_error($conn));
echo "failed";
exit();
}
What is the value in $return
after $result = mysqli_query($conn,$query);
$result = mysqli_query($conn,$query);
之后, $return
的值是多少? ? ?
For successful SELECT
, SHOW
, DESCRIBE
or EXPLAIN
queries, mysqli_query()
will return a mysqli_result object
. 对于成功的SELECT
, SHOW
, DESCRIBE
或EXPLAIN
查询, mysqli_query()
将返回mysqli_result object
。 For other successful queries mysqli_query()
will return TRUE
. 对于其他成功的查询, mysqli_query()
将返回TRUE
。 Returns FALSE
on failure. 失败时返回FALSE
。
So the value of $result
after your UPDATE
-query can only be true or false, nothing else. 因此, UPDATE
查询后的$result
值只能为true或false,不能有其他值。
Your echo..if...
can be simplified to one line: 您的echo..if...
可以简化为一行:
echo ($result?"success":"failed");
Hope this helps. 希望这可以帮助。
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