我有一个非常简单的用例,我正在检查表中是否存在某个值,但它似乎总是失败。这是我的 php 代码。
<?php
include "config.php";
$con = mysqli_connect($server_name,$mysql_user,$mysql_pass,$dbname);
if(!$con)
{
echo "Connection Error".mysqli_connect_error();
}
else{
//echo "";
}
$device_id = $_POST["device_id"];
$check = "SELECT magazine_id FROM registered_buyers WHERE device_id = $device_id";
$rs = mysqli_query($con,$check);
if(mysqli_num_rows($con,$rs) == 0)
{
$jsonarray = $_POST["jsonarray"];
echo "This will be inserted".$jsonarray;
}else
{
echo "User already registered";
}
?>
谁能指出我的错误。欢迎任何帮助或建议。谢谢。
最佳答案
您可以尝试遵循此代码。
<?php
include "config.php";
$con = mysqli_connect($server_name,$mysql_user,$mysql_pass,$dbname);
if(!$con){
echo "Connection Error".mysqli_connect_error();
}
$device_id = $_POST["device_id"];
$check = "SELECT magazine_id FROM registered_buyers WHERE device_id = ".$device_id;
$rs = mysqli_query($con, $check);
if(mysqli_num_rows($rs) == 0){
$jsonarray = $_POST["jsonarray"];
echo "This will be inserted".$jsonarray;
}else{
echo "User already registered";
}
?>
关于php - Mysql 检查 mysql 数据库中的值失败?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/38760412/