PHP警告:mysqli_fetch_assoc()期望恰好1个参数，其中3个给定 [英] PHP Warning: mysqli_fetch_assoc() expects exactly 1 parameter, 3 given in

问题描述

我收到此错误，我尝试过其他类似的线程，但没有帮助.

I get this error, I tried going through other similar threads but it didn't help.

这是php

$host = "http://www.domain.com";
$hostname = "localhost";
$username = "aaa";
$password = "sss";
$userstable = "ddd";
$dbName = "fff";

if ($url != $host){
$con = mysqli_connect($hostname, $username, $password, $dbName);
//@mysqli_select_db("$dbName");
// Check connection
if (mysqli_connect_errno($con))
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }
$query = mysqli_fetch_row(mysqli_query($con, "SELECT * FROM $userstable where(url = '$url')"));

if ($query == false){
$hits = "1";
$query2 = "INSERT INTO $userstable (url,hits) VALUES('$url','$hits')";
}
else {
$hitquery =  "SELECT `hits` FROM $userstable where url = '$url'";
$result = mysqli_query($con, $hitquery);
$hits = mysqli_fetch_assoc($result, 0,  'hits');
//$hits = mysqli_result(mysqli_query("SELECT `hits` FROM $userstable where url = '$url'"), 0,  "hits");
$query2 = "UPDATE $userstable SET `hits` = hits+1 where url = '$url'";
}
mysqli_query($con, $query2);
}
if(!$url) {
$url =  "$host";
}

或在第157行生成

$hits = mysqli_fetch_assoc($result, 0,  'hits');

如何修复此mysqli?

How to fix this mysqli ?

请帮助.

推荐答案

语法mysqli_fetch_assoc是:

Syntax mysqli_fetch_assoc is:

array mysqli_fetch_assoc ( mysqli_result $result )

尝试(对于PHP> = 5.4):

Try (for PHP >=5.4):

$hits = mysqli_fetch_assoc($result)['hits'];

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