dsg7513
2018-12-04 15:35
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如何通过jquery AJAX调用特定的php函数

I have a jquery ajax call to a php file. However, the file contains multiple defined functions, all somewhat similar.

$conn = mysqli_connect($servername, $username, $password, $dbname);


function loadall(){
    $querystring = "SELECT location, filetype, author FROM globalposts ORDER 
BY date DESC";
    $results = mysqli_query($conn, $querystring);

    $row = mysqli_fetch_array($results, MYSQLI_ASSOC);
    echo json_encode($row);
}

function jasperbriggsload(){
    $querystring = "SELECT title, location FROM omar ORDER BY date DESC";
    $results = mysqli_query($conn, $querystring);

    $row = mysqli_fetch_array($results, MYSQLI_ASSOC);
    echo json_encode($row);
}

I have not done overloaded methods with variables, though they might work with my db structure.

My main question is how to call a specific function in an external php through ajax. if I wanted to call loadall() or jasperbriggsload() through just ajax arguments

图片转代码服务由CSDN问答提供 功能建议

我有一个jquery ajax调用php文件。 但是，该文件包含多个已定义的函数，都有些类似。 

  $ conn = mysqli_connect（$ servername，$ username，$ password，$ dbname）; 
 
 \  nfunction loadall（）{
 $ querystring =“SELECT location，filetype，author FROM globalposts ORDER 
BY date DESC”; 
 $ results = mysqli_query（$ conn，$ querystring）; 
 
 $ row = mysqli_fetch_array（$ 结果，MYSQLI_ASSOC）; 
 echo json_encode（$ row）; 
} 
 
function jasperbriggsload（）{
 $ querystring =“SELECT title，location FROM omar ORDER BY date DESC”; 
 $ results = mysqli_query（  $ conn，$ querystring）; 
 
 $ row = mysqli_fetch_array（$ results，MYSQLI_ASSOC）; 
 echo json_encode（$ row）; 
} 
   
 
 
我没有使用变量重载方法，尽管它们可能适用于我的数据库结构。 
 
 
我的主要问题是如何通过ajax在外部php中调用特定函数。 如果我想通过ajax参数调用
 loadall（）
or 
 jasperbriggsload（）
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1条回答 默认 最新

  • douqian2524 2018-12-04 16:26
    已采纳

    You can use call_user_func passing the function name sent by ajax as argument. Like this:

    <?php
function test(){
    //your code
}

//This will run the test() function
call_user_func('test');

//This will run the function named as the content of $_REQUEST['funcName']
call_user_func($_REQUEST['funcName']);
?>
    已采纳该答案
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