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我在 php 函数内执行 mysql 查询时遇到了一些问题。我得到的错误是
Notice: Undefined variable: link in C:\path\api\inc\restFunctions.php on line 16
有几个文件互相调用，因此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;
}
}
serverList/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;
}
?>
serverList/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());
}
?>
我已验证调用的查询有效。我还通过使用查询数据库的该软件的不同页面验证了连接信息(上面隐藏的)是否有效。
我假设我一定在某处遗漏了引述或括号，但我对它可能在哪里感到困惑。