Question

我想在插入新记录之前检查是否已存在记录。但它到目前为止还没有工作，这是脚本：

<?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

尝试这样的事情：

<?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

试试这个：

//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错了。它只是

$checkexist->num_rows

或者您可以使用

mysqli_num_rows($checkexist)

希望这有帮助。

Answer 4

使用简单的INSERT IGNORE ...或INSERT ... ON DUPLICATE KEY UPDATE ...替换大部分代码。

后者允许您在记录已存在的情况下更改列（基于任何PRIMARY或UNIQUE键）。

mysqli_num_rows在PHP中没有使用MySQLi

4 个答案: