doudaotui4297 2018-04-12 03:29
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I was wondering how to construct the correct syntax for the if-else statement, or if there's something missing in my code.

<?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');
        } 
    }
?>
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2条回答 默认 最新

  • doubei8168 2018-04-12 03:55
    关注
    1. remove this if statment: if (!empty($docID)) {

    2. Make sure that u send with each post update the "docID" value

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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