Question

我试图将弹出窗口数据保存到database.but但是当我尝试在queryString中传递数据时抛出错误。它只是写“错误”。我不知道为什么它不调用update_clientstatus.php

Javscript：

function SaveNotes()
{
var notecontent = $("#txtPopNotes").val();
lookup(notecontent, globalNoteId);

overlay.appendTo(document.body).remove();
return false;
}

 function lookup(inputstr, eleid) {
    var inputString =  inputstr;

    // var row = parseFloat(eleId.substring(eleId.indexOf('_') + 1))+1;
    var noteId = '#noteContent_' + eleid;
    var editLinkId = '#editlink_' + eleid;
    var saveLinkId = '#savelink_' + eleid;
    var cancelLinkId = '#cancellink_' + eleid;
    var txtBoxId = '#txt_' + eleid;
    // var eleId = ele.id;

    $.post("update_clientstatus.php", {queryString: ""+inputString+"", id: ""+eleid+"" }, function(data){
        if(data=="success") {

            alert("Status updated successfully");
            $('.popup').hide();
            var noteId = '#noteContent_' + globalNoteId;
            $(noteId).html(inputstr);
            $('.message').html('Status Updated successfully');
        }
        else
        {
            **document.write('error');**
        }
    });

} // lookup

PHP代码（update_clientstatus.php）

 if(isset($_POST['queryString'])) {

        $queryString = $db->real_escape_string($_POST['queryString']);
        $eleid = $_POST['id'];          
        $query = mysqli_query($db,"UPDATE tblclientmaster SET status = '$queryString' WHERE id= '$eleid'");
}

Answer 1

试试这个

if(isset($_POST['queryString'])) {

        $queryString = $db->real_escape_string($_POST['queryString']);
        $eleid = $_POST['id'];          
        $query = mysqli_query($db,"UPDATE tblclientmaster SET status = '$queryString' WHERE id= '$eleid'");
        if ($query ) {
           echo "success";
        } else {
           echo "error";
        }
} else {
   echo "error";
}

将弹出窗口的数据保存到数据库中

1 个答案: