Question

我想为Android创建API。我正在使用php中的表单将数据插入MySQL，但是在插入数据之前，必须先在另一个表中检查该字段，然后再插入。如果该值存在于另一个表中，则将数据插入到主表中；如果不存在，则不插入数据。

这是我的php

<?php

if($_SERVER['REQUEST_METHOD']=='POST') {

$response = array();

$username = $_POST['username'];
$SN = $_POST['SN'];


require_once('dbConnect.php');

$sql = "SELECT * FROM produk WHERE SN ='$SN'";
$check = mysqli_fetch_array(mysqli_query($con,$sql));
if(isset($check)){
 $response["value"] = 0;
 $sql = "INSERT INTO farm (username,SN) VALUES('$username','$SN')";
 $response["message"] = "Successful";
 echo json_encode($response);
} else {
   $response["value"] = 1;
   $response["message"] = "oops! Try Again";
   echo json_encode($response);

}

mysqli_close($con);
} else {
$response["value"] = 0;
$response["message"] = "oops! Try Again";
echo json_encode($response);
}

?>

如何解决代码？

Answer 1

尝试下面的代码，您需要设置条件以检查另一个表中是否存在该值。检查mysqli_num_rows是否有其他表，如果num_rows> 0，则插入主表：

<?php

if($_SERVER['REQUEST_METHOD']=='POST') {

$response = array();

$username = $_POST['username'];
$SN = $_POST['SN'];


require_once('dbConnect.php');

$sql = "SELECT * FROM produk WHERE SN ='$SN'";
$check = mysqli_num_rows(mysqli_query($con,$sql));
if($check > 0){
 $response["value"] = 0;
 $sql = "INSERT INTO farm (username,SN) VALUES('$username','$SN')";
 $response["message"] = "Successful";
 echo json_encode($response);
} else {
   $response["value"] = 1;
   $response["message"] = "oops! Try Again";
   echo json_encode($response);

}

mysqli_close($con);
} else {
$response["value"] = 0;
$response["message"] = "oops! Try Again";
echo json_encode($response);
}

?>

如何使API PHP检查该值是否存在于另一个表中然后插入？

1 个答案: