Question

我想知道如何为if-else语句构造正确的语法，或者我的代码中是否缺少某些内容。

<?php
    include "../dbcon.php";
    session_start();
    ob_start();
    $sql = mysqli_query($con,"SELECT * FROM clientdocuments  WHERE docID = $_POST[docID]");
    $rows = mysqli_fetch_array($sql, MYSQLI_ASSOC);
    //IF CSS input value is filled
    if(!empty($_POST)){
        $output = '';  
        $message = '';  
        $docID = mysqli_real_escape_string($con, $_POST["docID"]);
        $docSIG_Contract = mysqli_real_escape_string($con, $_POST["docSIG_Contract"]); 
        //I don't get what this "if(isset($_POST["docID"])){" purpose (Sorry very new to php)
        if(isset($_POST["docID"])){
            if (!empty($docID)) {
                $query = "UPDATE clientdocuments(docID, docSIG_Contract) VALUES('$docID', '$docSIG_Contract');"; //UPDATE ONCE docID ALREADY EXIST ON THE DATABASE 
            } else {
                $query = "INSERT INTO clientdocuments(docID, docSIG_Contract) VALUES('$docID', '$docSIG_Contract');"; //INSERT IF THE docID doesn't exist yet
            }
            $str = mysqli_query($con,$query);
            if(!$str){
                echo 'FAILED';
            }
        }else{
            header('HTTP/1.1 500 Internal Server Booboo');
            header('Content-Type: application/json; charset=UTF-8');
        } 
    }
?>

Answer 1

删除此if语句：if (!empty($docID)) {
确保您随每个帖子发送更新＆＃34; docID＆＃34;值

Answer 2

if（isset（$ _ POST [＆＃34; docID＆＃34;]））语句检查名称为docID的输入是否具有值。

if（！empty（$ _ POST））我不确定这是否有效，我的猜测是你试图检查请求方法是否是POST（如果点击了保存按钮））。为此，我使用 if（$ _SERVER [＆＃39; REQUEST_METHOD＆＃39;] ===＆＃39; POST＆＃39;）{

然后我会检查docID是否有值ie （isset（$ _ POST [＆＃34; docID＆＃34;]））或（！empty（$ _ POST [＆＃34; docID＆＃34;]）） isset和！之间的区别 What's the difference between 'isset()' and '!empty()' in PHP?

如果有值，$ query将是update语句如果没有值$ query将是insert语句在这种情况下不要输入DocID值（因为它总是0，这也会导致错误）

希望有意义！

保存按钮，一旦记录已保存在数据库中，也可以更新

2 个答案: