Mysql_num_row（）期望參數1是資源

Question

我已試圖檢查sql數據庫中存在的行，但它總是返回此錯誤。我試圖研究他們提供了一些解決方案，但不工作。請幫忙！！

$usersexist = mysqli_query($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

你是mixin mysqi和mysql。試試這個 -

$usersexist = mysqli_query($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。您應該在這裏使用$ userexist-> num_rows（object context）或mysqli_num_rows（$ userexist）過程上下文。你也應該檢查$ userexist是否只返回false（失敗的查詢）。所以像這樣：

$usersexist = mysqli_query($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，則無法使用mysql。

mysql_num_rows($usersexist) 

正確

mysqli_num_rows($usersexist)