我正在尝试打印/回显以前文件中的值，但是没有显示错误，而是在空白页上显示的所有内容都是“未定义”。我已经研究并尝试了几种方法，但没有任何效果。请帮忙。
<?PHP
$user_name = "root";
$password = "";
$database = "leadership_program";
$server = "localhost";
$db_handle = mysql_connect($server, $user_name, $password);
$db_found = mysql_select_db($database, $db_handle);
if ($db_found) {
if (isset($_POST['survey_id'])) {
$survey_id = $_POST['survey_id'];
echo $survey_id;
}
if (isset($_POST['marks'])) {
foreach ($_POST['marks'] as $value) {
echo"$value";
}
}
if (isset($_POST['id'])) {
$id = $_POST['id'];
echo $id;
}
// $SQL2 = "UPDATE answer_table SET marks='$value' WHERE survey_id= '$survey_id' AND student_id= '$id'";
//$result2 = mysql_query($SQL2);
//mysql_close($db_handle);
} else {
print "Database NOT Found ";
mysql_close($db_handle);
// header("Location: surveyView.php");
}
?>
这里是displayresult.php
<form action="student_mark_save.php" method="POST"> //<?php...more codes here
if ($strucrow["qns$i"] === 'radio' || $strucrow["qns$i"] === 'checkbox') {
foreach ($arr as $b) {
echo "<br/>";
if (strpos($b, '%#%') !== false) {
$c = substr($b, 3, -2);
//echo $c;
$d = str_replace("$arr[0] :-", ':', $c);
echo $d, "<br/>";
echo "<br/>";
echo "<tr> Marks : <input type=\"text\" name=\"marks[]\"></tr><br />";
//echo $b;
} else {
echo $b;
// echo "is not with comment qns";
}
}
} else if ($strucrow["qns$i"] === 'comment') {
foreach ($arr as $b) {
echo $b;
echo "<tr> Marks : <input type=\"text\" name=\"marks[]\"></tr><br />";
// echo "is not with comment qns";
}
} else {
}
echo "<p/>";
}
$marksquery = sprintf(
"SELECT marks FROM answer_table WHERE survey_id = '%d' AND student_id = '$studentid' ", mysql_real_escape_string($survey_id)
);
$marksQuer = mysql_query($marksquery) or die(mysql_error());
$marksrow = mysql_fetch_assoc($marksQuer);
echo "<td><input type=\"hidden\" value= \"$survey_id\" name=\"survey_id\"></td><br />" ;
echo "<td><input type=\"hidden\" value= \"$studentid\" name=\"id\"></td><br />" ;
echo "<p><input type=\"submit\" value=\"Update\"></p>";
?>
</form>
最佳答案
对于未定义的意思，全局$_POST[]中不存在某些东西
检查尝试在您的第一个 if 语句中打印 $_POST 为:
if ($db_found) {
print_r($_POST);
exit();
if (isset($_POST['survey_id'])) {
$survey_id = $_POST['survey_id'];
echo $survey_id;
}
现在看看你在$_POST中有什么，它可能对你有帮助...谢谢
关于PHP:空白页上显示未定义，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/20922957/