Question

我试过检查sql数据库中的存在行，但它总是返回此错误。我试过研究他们给了一些解决方案，但没有工作。请帮忙!!

$usersexist = mysqli_query($con, "SELECT phone FROM usersacc WHERE phone = '$pho'");

        $count= mysql_num_rows($usersexist) or die(mysql_error());
        if($count > 0)
        {
         $result_data = array( 
           'ResultArray' => 'Exist',
            ); 
        }
        else
        {
        mysqli_query($con,"INSERT INTO usersacc
(phone, password) VALUES('$pho', '$pass')");


        #Build the result array (Assign keys to the values) 
        $result_data = array( 
           'ResultArray' => 'success',
            ); 
        }
        #Output the JSON data 
        echo json_encode($result_data);

Answer 1

您正在将mysql与mysql混合使用。试试这个 -

$usersexist = mysqli_query($con, "SELECT phone FROM usersacc WHERE phone = '$pho'") or die(mysqli_error());

$count= mysqli_num_rows($usersexist);

if($count > 0) {
//rest of your code

Answer 2

Mysql不是mysqli。您应该在这里使用$ userexist-＆gt; num_rows（对象上下文）或mysqli_num_rows（$ userexist）程序上下文。另外你应该检查$ userexist是否只返回false（查询失败）。像这样：

$usersexist = mysqli_query($con, "SELECT phone FROM usersacc WHERE phone = '$pho'");

if(!$userexist)
    die("query failed");


$count = mysqli_num_rows($userexist);
if($count > 0)
{
 $result_data = array( 
   'ResultArray' => 'Exist',
    ); 
}
else
{
mysqli_query($con,"INSERT INTO usersacc
(phone, password) VALUES('$pho', '$pass')");


#Build the result array (Assign keys to the values) 
$result_data = array( 
   'ResultArray' => 'success',
    ); 
}
#Output the JSON data 
echo json_encode($result_data);

Answer 3

如果您使用mysqli，则无法使用mysql。

mysql_num_rows($usersexist)

正确

mysqli_num_rows($usersexist)

Mysql_num_rows（）期望参数1是资源

3 个答案: