【发布时间】:2014-09-24 19:33:49
【问题描述】:
几周前我问这是另一个论坛,没有人能够回答它,希望这里有人能回答。选择的状态进入数据库,但是当页面刷新时,它只显示选择而不是状态。性别也是如此,但性别甚至没有在数据库中发布。
<?php
if( !isset( $_SESSION ) ){
session_start();
}
$con=mysqli_connect("localhost", "root", "", "test");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
// escape variables for security
$firstname = mysqli_real_escape_string($con, $_POST['firstname']);
$lastname = mysqli_real_escape_string($con, $_POST['lastname']);
$gender = mysqli_real_escape_string($con, $_POST['gender']);
$number = mysqli_real_escape_string($con, $_POST['number']);
$email = mysqli_real_escape_string($con, $_POST['email']);
$addressone = mysqli_real_escape_string($con, $_POST['addressone']);
$addresstwo = mysqli_real_escape_string($con, $_POST['addresstwo']);
$city = mysqli_real_escape_string($con, $_POST['city']);
$state = mysqli_real_escape_string($con, $_POST['state']);
$zip = mysqli_real_escape_string($con, $_POST['zip']);
$sql="UPDATE users SET firstname='$firstname', lastname = '$lastname', gender = '$gender,'number='$number', email = '$email',addressone='$addressone', addresstwo= '$addresstwo',`city`='$city', state = '$state', zip = '$zip' WHERE id='" .$_SESSION['id']."'";
if (!mysqli_query($con,$sql)) {
die('Error: ' . mysqli_error($con));
}
header("Location:dashboard.php");
mysqli_close($con);
?>
Part of the entire form
<div class="form-group">
<label class="col-md-5 control-label" name="state">State</label>
<div class="col-md-2">
<select id="state" name="state" class="form-control" value="<?php echo $state; ?>">
<option>Select</option>
<option>Alabama</option>
<option>Alaska</option>
<option>Arizona</option>
<option>Arkansas</option>
<option>California</option>
<option>Colorado</option>
<option>Connecticut</option>
<option>Delaware</option>
<option>District of Colombia</option>
<option>Florida</option>
<option>Georgia</option>
<option>Hawaii</option>
<option>Idaho</option>
<option>Illinois</option>
<option>Indiana</option>
<option>Iowa</option>
<option>Kansas</option>
<option>Kentucky</option>
<option>Louisiana</option>
<option>Maine</option>
<option>Maryland</option>
<option>Massachusetts</option>
<option>Michigan</option>
<option>Minnesota</option>
<option>Mississippi</option>
<option>Missouri</option>
<option>Montana</option>
<option>Nebraska</option>
<option>Nevada</option>
<option>New Hampshire</option>
<option>New Jersey</option>
<option>New Mexico</option>
<option>New York</option>
<option>North Carolina</option>
<option>North Dakota</option>
<option>Ohio</option>
<option>Oklahoma</option>
<option>Oregon</option>
<option>Pennsylvania</option>
<option>Rhode Island</option>
<option>South Carolina</option>
<option>South Dakota</option>
<option>Tennessee</option>
<option>Texas</option>
<option>Utah</option>
<option>Vermont</option>
<option>Washington</option>
<option>West Virginia</option>
<option>Wisconsin</option>
<option>Wyoming</option>
<option>American Samoa</option>
<option>Federated States of Micronesia</option>
<option>Guam</option>
<option>Marshall Islands</option>
<option>Northern Mariana Islands</option>
<option>Puerto Rico</option>
<option>Virgin Islands</option>
<option>Palau</option>
<option>AA</option>
<option>AE</option>
<option>AP</option>
</select>
</div>
</div>
Gender portion of form
<!-- Multiple Radios (inline) -->
<div class="form-group">
<label class="col-md-5 control-label" for="gender">Gender</label>
<div class="col-md-2">
<label class="radio-inline" for="gender-0">
<input type="radio" name="gender" id="gender-0"checked="checked" value="<?php echo $gender; ?>">
Male
</label>
<label class="radio-inline" for="gender-1">
<input type="radio" name="gender" id="gender-1" value="<?php echo $gender; ?>">
Female
</label>
</div>
</div>
【问题讨论】:
标签: php mysql mysqli phpmyadmin