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我一直在教授自己過去3個月的代碼，並且正在研究一個超出我的知識水平的項目。因此，在我們創建的網站上，我們有一個用戶創建配置文件的頁面，任何訪問該網站的人都可以看到該信息，他們輸入的信息將存儲在數據庫表中。我想要做的只是顯示該特定用戶的信息。現在我創建了三個不同的用戶，當我登錄到每個用戶時，它顯示每個用戶的相同信息。我相信這個問題已經得到解答，我只是不知道該怎麼說。以下是我目前使用的輸入表單和.php文件。如何輸出特定於用戶的數據庫中的信息
<div class="modal-body">
<form action="edit_profile.php" method="post">
<input type="text" class="form-control" id="exampleInputEmail1" placeholder="Name" name="name" required="">
<br>
<textarea class="form-control" name="about_me" rows="3" placeholder="About Me (300 Characters Max)"></textarea></textarea>
<br>
<input type="text" class="form-control" id="exampleInputEmail" placeholder="My Specialties" name="specialty" required="">
<br>
<input type="text" class="form-control" id="exampleInputEmail" placeholder="City" name="city" required="">
<br>
<input type="text" class="form-control" id="exampleInputEmail" placeholder="State" name="state" required="">
<br>
<input type="email" class="form-control" id="exampleInputEmail" placeholder="Email" name="email" required="">
<br>
<input type="text" class="form-control" id="exampleInputEmail" placeholder="Website" name="website" required="">
<br>
<input type="email" class="form-control" id="exampleInputEmail" name="facebook" placeholder="Facebook Link">
<br>
<input type="email" class="form-control" id="exampleInputEmail" name="instagram" placeholder="Instagram Link">
<br>
<input type="email" class="form-control" id="exampleInputEmail" name="twitter" placeholder="Twitter Link">
<br>
<input type="email" class="form-control" id="exampleInputEmail" name="google" placeholder="Google+ Link">
<br>
<input type="submit" class="btn btn-primary" value="submit">
<button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
</form>
</div>
</div>
</div>
</div>
<!-- End Edit Profile Modal -->
<h4><strong>About Me: </strong></h4>
<?php echo $row['about_me'];?>
<h4><strong>My Specialties: </strong></h4>
<?php echo $row['specialty'];?>
<h4><strong>Location: </strong> </h4>
<?php echo $row['city'];?>, <?php echo $row['state'];?>
<h4><strong>Get Connected: </strong></h4>
<h5><strong>Email:</strong> <?php echo $row['email'];?> </h5>
<h5><strong>Website:</strong> <?php echo $row['website'];?></h5>
<h5><strong>Facebook:</strong> <?php echo $row['facebook'];?></h5>
//DATABASE CONNECT
<?php
$host="localhost";
$username="XXXXXXX";
$password="XXXXXXX";
$db_name="photographer_directory";
$tbl_name="qls3_profile";
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$result = mysql_query("SELECT * FROM qls3_profile");
$row = mysql_fetch_array($result);
?>
<?php
include_once('includes/db_connect.php');
$name=$_POST['name'];
$about_me=$_POST['about_me'];
$specialty=$_POST['specialty'];
$city=$_POST['city'];
$state=$_POST['state'];
$email=$_POST['email'];
$website=$_POST['website'];
$facebook=$_POST['facebook'];
$instagram=$_POST['instagram'];
$twitter=$_POST['twitter'];
$google=$_POST['google'];
// Insert data into mysql
$sql="INSERT INTO $tbl_name(name, about_me, specialty, city, state, email, website, facebook, instagram, twitter, google)VALUES('$name', '$about_me', '$specialty', '$city', '$state', '$email', '$website', '$facebook', '$instagram', '$twitter', '$google')";
$result=mysql_query($sql);
// if successfully insert data into database, displays message "Successful".
if($result){
echo "Successful";
echo "<BR>";
echo "<a href='insert.php'>Back to main page</a>";
}
else {
echo "ERROR";
}
header('Location: userprofile.php') ;
?>
<?php
// close connection
mysql_close();
?>
使用['WHERE']（http://www.sql-tutorial.net/SQL-WHERE.asp）子句。 –
Where子句不適用於我正在嘗試完成的任務。基本上用戶登錄並創建一個配置文件，我需要這些信息顯示在他們的配置文件中。我正在尋找遍佈這個網站，並找不到任何將與我正在做的工作。 – user3587862