在php函数中执行mysql查询时遇到一些问题。我得到的错误是
Notice: Undefined variable: link in C:'path'api'inc'restFunctions.php on line 16
有几个文件相互调用,所以我将尝试概述必要的信息。
访问的URL:
localhost/serverList/api/rest.php?action=allServers
serverList/api/rest.php
<?php
include 'inc/restFunctions.php';
$possibleCalls = array('allServers','allEnvs','allTypes','false');
if(isset($_GET['action'])){
$action = $_GET['action'];
}
else{
$action = 'false';
}
if(in_array($action,$possibleCalls)){
switch ($action){
case 'allServers':
$return = allServers();
break;
case 'allEnvs':
$return = allEnvs();
break;
case 'allTypes':
$return = allTypes();
break;
case 'false':
$return = falseReturn();
break;
}
}
服务器列表/api/inc/restFunctions.php
<?php
include ('inc/config.php');
function allServers(){
$serverInfoQuery = "SELECT * FROM servers"
$allServerResults = $link->query($serverInfoQuery);
$json = array();
while($row = $allServerResults->fetch_assoc()){
$json[]['serverID'] = $row['serverID'];
$json[]['environment'] = $row['environment'];
$json[]['type'] = $row['type'];
$json[]['serverIP'] = $row['serverIP'];
$json[]['serverDescription'] = $row['serverDescription'];
$json[]['serverCreatedBy'] = $row['serverCreatedBy'];
$json[]['serverCreatedDtTm'] = $row['serverCreatedDtTm'];
$json[]['serverUpdatedBy'] = $row['serverUpdatedBy'];
$json[]['serverUpdatedDtTm'] = $row['serverUpdatedDtTm'];
}
$jsonResults = json_encode($json);
return $jsonResults;
}
?>
服务器列表/api/inc/config.php
<?php
$host = 'localhost';
$user = 'userName';
$password = 'password';
$database = 'database';
$link = new mysqli($host, $user, $password, $database);
if (mysqli_connect_errno()) {
exit('Connect failed: '. mysqli_connect_error());
}
?>
我已经验证了被调用的查询是否有效。我还验证了连接信息(上面屏蔽的)是通过使用该软件的另一个查询数据库的页面来工作的。
我想我一定在某个地方遗漏了一个引号或paren,但我不知道它可能在哪里。
问题出在PHP变量作用域上。在首次引用$link变量之前,在allServer()函数中添加此行:
global $link;
点击此处查看更多信息:http://php.net/manual/en/language.variables.scope.php
在我看来,使用全局变量不是一个好的解决方案。您可能会意外地在某个范围内覆盖$link($link是您可能用于其他目的的变量的常用名称)变量,从而导致大量混乱和调试困难。只需将其作为函数参数传递-更干净、更容易阅读:
function AllServers($link) {
$serverInfoQuery = "SELECT * FROM servers";
$allServerResults = $link->query($serverInfoQuery);
//More PHP code
}
if(in_array($action,$possibleCalls)){
switch ($action){
case 'allServers':
$return = allServers($link);
break;
}
}
老实说,更好的解决方案是使用一些通用类/函数来建立mysql连接,如下所示:
class DB {
private static $link = null;
public static function getConnectionResource() {
//In that way we "cache" our $link variable so that creating new connection
//for each function call won't be necessary
if (self::$link === null) {
//Define your connection parameter here
self::$link = new mysqli($host, $user, $password, $database);
}
return self::$link;
}
}
function getAllServers() {
$link = DB::getConnectionResource();
//Preform your query and return results
}
使用全局变量
function allServers(){
global $link
...
...
...
... your code follows