我使用以下代码得到以下错误：

注意：未定义的变量：第35行的C：\ xampp \ htdocs \ test \ projects \ Learning \ php \ Databases \ Forms和Databases \ project.php中的错误

代码：

$clicked = false;

if($clicked == false && isset($_POST['submit'])) {

if ($_POST['label'] == '') {

echo "

You must enter in a label!

";

$error = true;

}

if ($_POST['url'] == '') {

echo "

You must enter in a url!

";

$error = true;

}

if ($_POST['status'] == '') {

echo "

You must enter in a status (0 or 1)!

";

$error = true;

}

}

if ($error != true) {

if (isset($_POST['submit']) && $_POST['submit'] == '1' ) {

$query = "INSERT INTO nav (label, url, target, status, position) VALUES ('$_POST[label]', '$_POST[url]', '$_POST[target]', $_POST[status], $_POST[position])";

$result = mysqli_query($dbc, $query);

if ($result) {

echo "

You've added a new navigation link!

";

}

else {

echo "

An error has occured!

";

echo mysqli_error($dbc);

echo '

'.$query.'

';

}

$clicked = true;//submit button replaced

}

}

?>

Navigation

*Required Fields

*Label:

Position:

*URL:

Target:

*Status:

if ($clicked == false) {

echo '

Add Navigation Link

';

}

else {

echo '

Do Another

';

}

?>

当我做出以下更改时：

if($clicked == false && $_POST['submit'] == "1") {

if ($_POST['label'] == '') {

echo "

You must enter in a label!

";

$error = true;

}

if ($_POST['url'] == '') {

echo "

You must enter in a url!

";

$error = true;

}

if ($_POST['status'] == '') {

echo "

You must enter in a status (0 or 1)!

";

$error = true;

}

}

我收到这些错误：

注意：未定义的索引：在第18行的C：\ xampp \ htdocs \ test \ projects \ Learning \ php \ Databases \ Forms和Databases \ project.php中提交

＆

注意：未定义的变量：第35行的C：\ xampp \ htdocs \ test \ projects \ Learning \ php \ Databases \ Forms和Databases \ project.php中的错误

很明显，名称“提交”的按钮不会因任何原因被“看到”;我相信 . 这对我来说有点意义，有点......如果我没有弄错：php从头到尾以线性方式读取，并且因为表单在if语句之下，所以索引还不存在 . 我认为这进一步证实了这样一个事实：一旦我单击提交按钮，所有错误都会消失，因此执行if语句中的if语句的错误(echo)语句 .

这令人难以置信 . 这也不起作用......

if(isset($_POST['submit']) && $_POST['submit'] == '1') {

if (isset($_POST['label']) && $_POST['label'] == '') {

echo "

You must enter in a label!

";

$error = true;

}

if (isset($_POST['url']) && $_POST['url'] == '') {

echo "

You must enter in a url!

";

$error = true;

}

if (isset($_POST['status']) && $_POST['status'] == '') {

echo "

You must enter in a status (0 or 1)!

";

$error = true;

}

}

...在以前版本的代码中，isset和"is equal to"的组合用于if语句's conditions, solved the Unidentified index problem, as it pertained to the $_POST[' submit '], here' s代码：ps：因为它属于这个特定的代码块，以下链接的伙伴tutorial尽管我和他做了完全相同的事情，但我正在跟随它，并没有出现任何这些错误 .

$clicked = false;

if (isset($_POST['submit']) && $_POST['submit'] == '1' ) {

$query = "INSERT INTO nav (label, url, target, status, position) VALUES ('$_POST[label]', '$_POST[url]', '$_POST[target]', $_POST[status], $_POST[position])";

$result = mysqli_query($dbc, $query);

if ($result) {

echo "

You've added a new navigation link!

";}

else {

echo "

An error has occured!

";

echo mysqli_error($dbc);

echo '

'.$query.'

';

}

$clicked = true;//submit button replaced

}

?>

Navigation

*Required Fields

*Label:

Position:

*URL:

Target:

*Status:

if ($clicked == false) {

echo '

Add Navigation Link

';

}

else {

echo '

Do Another

';

}

?>

再次，这工作得很好，没有错误 . 那么为什么我在我发布的第一个代码块中得到了undefine变量错误？未定义的变量是由于后续的if语句未能执行而导致的，这就是我假设与索引问题有关的问题，但是，错误并没有反映出来！

当我用$ clicked == false替换条件时，如下所示：

$clicked = false;

if($clicked == false) {

if ($_POST['label'] == '') {

echo "

You must enter in a label!

";

$error = true;

}

if ($_POST['url'] == '') {

echo "

You must enter in a url!

";

$error = true;

}

if ($_POST['status'] == '') {

echo "

You must enter in a status (0 or 1)!

";

$error = true;

}

}

我得到了这三个未定义的索引错误ALONG WITH BLOODY CODE，它显然成功执行，尽管三个索引被认为是未定义的：

注意：未定义的索引：第20行的C：\ xampp \ htdocs \ test \ projects \ Learning \ php \ Databases \ Forms和Databases \ project4.php中的标签您必须输入标签！

注意：未定义的索引：第24行的C：\ xampp \ htdocs \ test \ projects \ Learning \ php \ Databases \ Forms和Databases \ project4.php中的url你必须输入一个url！

注意：未定义的索引：第28行的C：\ xampp \ htdocs \ test \ projects \ Learning \ php \ Databases \ Forms和Databases \ project4.php中的状态您必须输入状态(0或1)！