Php Mysql数据库中的Mysql值检查失败?
我有一个非常简单的用例,在这个用例中,我检查表中是否存在某个值,它似乎总是失败Php Mysql数据库中的Mysql值检查失败?,php,mysql,Php,Mysql,我有一个非常简单的用例,在这个用例中,我检查表中是否存在某个值,它似乎总是失败 <?php include "config.php"; $con = mysqli_connect($server_name,$mysql_user,$mysql_pass,$dbname); if(!$con) { echo "Con
<?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";
}
?>
任何人都可以指出我的错误。欢迎任何帮助或建议。谢谢。 < P>因为我没有足够的评论来添加评论,我会考虑<代码> DEVICIOSIDID/COD>是字符串,如果是这样的话,试试这样的:
"SELECT magazine_id FROM registered_buyers WHERE device_id = '$device_id'";
添加
”
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";
}
?>
了解防止SQL注入的准备语句执行SQL语句后检查错误没有错误是的,我肯定会这样做,但你能指出它不起作用的原因吗。@Jentry remove$con
frommysqli_num_rows()
-仅使用mysqli_num_rows($rs)
Source:@Dave刚刚这么做了,它仍然给出相同的结果。