php - 在循环 php 中命中数据库

Question

我是 PHP 的新手。我只是创建一个简单的页面，从数据库中获取数据并生成 XML。有两个表，一个有竞争对 watch ，第二个有这些竞争对手的排名，所以我能够获取候选数据，当我循环到该数据并尝试在循环中再次访问数据库时，出现以下错误:

Undefined variable: mysqli in

Fatal error: Call to a member function query() on null in

我尝试了一些方法，但都没有用。这是我的代码:

<?php
/** create XML file */ 
$mysqli = new mysqli("localhost", "root", "******", "****");
/* check connection */
if ($mysqli->connect_errno) {
   echo "Connect failed ".$mysqli->connect_error;
   exit();
}else{
}
$query = "select * from competitors where eventid=290 order by 1 desc LIMIT 0, 10;";
$booksArray = array();
if ($result = $mysqli->query($query)) {
/* fetch associative array */
   while ($row = $result->fetch_assoc()) {
   array_push($booksArray, $row);
  }
   if(count($booksArray)){
createXMLfile($booksArray);

 }
/* free result set */
   $result->free();
}
function createXMLfile($booksArray){
   $filePath = 'book.xml';
   $dom     = new DOMDocument('1.0', 'utf-8'); 
   $root      = $dom->createElement('books'); 
   for($i=0; $i<count($booksArray); $i++){
$eventid        =  $booksArray[$i]['eventid'];  
 $fee      =  $booksArray[$i]['fee']; 
$competitorid_system    =  $booksArray[$i]['competitorid'];
$datetimeentered     =  $booksArray[$i]['datemodified']; 
// $bookISBN      =  $booksArray[$i]['ISBN']; 
//  $bookCategory  =  $booksArray[$i]['category'];  
$book = $dom->createElement('book');
 $book->setAttribute('eventid', $eventid);
$name     = $dom->createElement('fee', $fee); 
$book->appendChild($name); 
$author   = $dom->createElement('competitorid_system', $competitorid_system); 

 $book->appendChild($author); 

 $price    = $dom->createElement('datetimeentered', $datetimeentered); 

$book->appendChild($price); 

// fetch other data
$query = "select * from ranking where eventid=290 and competitorid=".$competitorid_system;;

 echo "Query".$query;
   //exit();
$booksRankingArray = array();
if ($result = $mysqli->query($query)) {
/* fetch associative array */
while ($row = $result->fetch_assoc()) {
array_push($booksRankingArray, $row);
}
if(count($booksRankingArray)){
createXMLfile($booksRankingArray);
 }
/* free result set */
$result->free();
}
$root->appendChild($book);
}
$dom->appendChild($root); 
$dom->save($filePath); 
} 
/* close connection */
$mysqli->close();

Answer 1

$mysqli 变量超出了 createXMLfile() 的范围。

The scope of a variable is the context within which it is defined. For the most part all PHP variables only have a single scope. This single scope spans included and required files as well.

Please refer to the PHP Manual to understand the variable scope .

为了解决问题，将函数签名更改为

function createXMLfile($mysqli, $booksArray)
{ 
    // … rest of code …
}

然后将变量从外部作用域传递到函数作用域:

$mysqli = new mysqli("localhost", "root", "******", "****");
$booksArray = array();    
createXMLfile($mysqli, $booksArray);

您还可以使用全局变量从全局范围中提取变量:

function createXMLfile($booksArray)
{ 
    global $mysqli;
    // … rest of code …    
}

But using global variables is generally discouraged因为它会导致紧密耦合的代码并使代码不那么容易推理。