如何解决提交表单时Mysqli表未更新如何解决?
这是无法更新数据库表的代码:
mysqli_query($connect,"UPDATE questions1 SET votes='$newvotes' WHERE rateid='$rateid' AND question1='$rateoption'");
mysqli_query($connect,"UPDATE rates1 SET ipaddress='$newipaddress' WHERE rateid='$rateid'");
这是我的完整代码:
<?php
$rateid = $_GET['rateid'];
$connect = mysqli_connect('localhost','root','','rate');
$query = "SELECT * FROM rates1 WHERE rateid='$rateid'";
$q = mysqli_query($connect,$query);
while($row = mysqli_fetch_array($q)) {
$id = $row[0];
$title = $row[1];
$rateid = $row[2];
$ipaddress = $row[3];
echo "<h1 id = 'lmk'>$title</h1><br>";
?>
<table>
<form action="" method="POST">
<?php
$questions = "SELECT * FROM questions1 WHERE rateid='$rateid'";
$q2 = mysqli_query($connect,$questions);
while($r = mysqli_fetch_array($q2)) {
$question = $r[1];
$votes = $r[2];
$newvotes = $votes + 1;
$ip = $_SERVER['REMOTE_ADDR'];
$newipaddress = $ipaddress."$ip,";
if (isset($_POST['rate'])) {
$rateoption = $_POST['rateoption'];
if ($rateoption == "") {
die('<div class= "blm">' . "You didn't select an option." . '<br><br>' . '</div>');
} else {
$ipaddresse = explode(",",$ipaddress);
if (in_array($ip,$ipaddresse)) {
die('<div class= "blm">' . "You've already rated" . '<br><br>' . '</div>');
} else {
mysqli_query($connect,"UPDATE rates1 SET ipaddress='$newipaddress' WHERE rateid='$rateid'");
die('<div class= "blm">' . "Your rating was casted successfully" . '<br><br>' . '</div>');
}
}
}**
'<div class = "mnbh">';
echo '<tr><td><div class ="loij">'.$question.'</div></td><td><div class = "lopk"><input type="radio" name="rateoption" value="'.$question.'" /><div class = "mnbv"></div></div></td></tr>';
}
}
?>
<tr><td><br><br><div class="cogat"><input type = "submit" name = "rate" value = "Submit"></div></td></tr>
</div>
</div>
</div>
</form></table>
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