mysqli_num_rows在MySQLi和PHP中不起作用

Question

我想檢查一下是否存在記錄，然后再插入新記錄。 但這到目前為止還行不通，這是腳本：

<?php
session_start();
$uname = $_SESSION['username'];
$friend = $_GET["newfriend"];

$db = new mysqli("localhost", "...", "....", "...");
if($db->connect_errno > 0){
    die("Unable to connect to database: " . $db->connect_error);
}

$checkexist = $db->query("SELECT * FROM friends WHERE (username_own='$uname', username='$friend')");

//create a prepared statement
$stmt = $db->prepare("INSERT INTO friends (`username_own`, `username`) VALUES (?,?)");

//bind the username and message
$stmt->bind_param('ss', $uname, $friend);

if ($checkexist->mysqli_num_rows == 0) {
  //run the query to insert the row
  $stmt->execute();
}

Answer 1

嘗試這樣的事情：

<?php

/* Check if user exists */
$query = "SELECT count(1) FROM friends WHERE username_own=? AND username=?";

if($stmt = $db->prepare($query)){
  $stmt->bind_param('ss', $uname, $friend);
  $stmt->execute();
  $stmt->bind_result($count_rows);
  $stmt->fetch();
  $stmt->close();
}else die("Failed to prepare");

/* If user doesn't exists, insert */
if($count_row == 0){

    $query = "INSERT INTO friends (`username_own`, `username`) VALUES (?,?)";

    if($stmt = $db->prepare($query)){
      $stmt->bind_param('ss', $uname, $friend);
      $stmt->execute();
      $stmt->close();
    }else die("Failed to prepare!");
}

Answer 2

嘗試這個：

//create a prepared statement
$stmt = $db->prepare("INSERT INTO friends (`username_own`, `username`) VALUES (?,?)");

//bind the username and message
$stmt->bind_param('ss', $uname, $friend);

if ($checkexist->mysqli_num_rows == 0 || $checkexist->mysqli_num_rows <> 0) {
  //run the query to insert the row
  $stmt->execute();
}

Answer 3

$checkexist->mysqli_num_rows錯誤。 只是

$checkexist->num_rows

或者你可以使用

mysqli_num_rows($checkexist)

希望這可以幫助。

Answer 4

用簡單的INSERT IGNORE ...或INSERT ... ON DUPLICATE KEY UPDATE ...替換大多數代碼。

如果記錄已經存在（基於任何PRIMARY或UNIQUE鍵），則后者使您可以更改列。