dongmei9961 2017-01-04 02:28
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已采纳

变量在If语句中不起作用

As I posted here, I managed to get 2 variable that I would like to use for further processing. Unfortunatly, I can't use is inside that if statement.

if (isset($_POST['modifybtn'])) { //Now we have our variables all set
    $SubmissionID = $_POST['modifybtn'];
    $CourseID = $_POST['courseid'];
    }
//We do some irrelevant stuff here.... and then

echo $CourseID; //Works
if (isset($_POST['savebtn'])) {
        echo $CourseID; //Not Working

    saveData($AppID, $CourseID);
    header("Location: applicantCase.php");
}
if (isset($_POST['nextbtn'])) {
    saveData($AppID, $CourseID);
    header("Location: ApplicantApplyEducation.php");
}

Why is that happening, one line before was working?

enter image description here

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1条回答 默认 最新

  • douzhang1955 2017-01-04 04:08
    关注

    First, You can define outside the if condition.

     $SubmissionID = "";
     $CourseID = "";
    if (isset($_POST['modifybtn'])) { //Now we have our variables all set
        $SubmissionID = $_POST['modifybtn'];
        $CourseID = $_POST['courseid'];
        }
    //We do some irrelevant stuff here.... and then
    
    echo $CourseID; //Works
    if (isset($_POST['savebtn'])) {
            echo $CourseID; //Now it will Working
    
        saveData($AppID, $CourseID);
        header("Location: applicantCase.php");
    }
    if (isset($_POST['nextbtn'])) {
        saveData($AppID, $CourseID); // Now it will work
        header("Location: ApplicantApplyEducation.php");
    }
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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