[英]Only the first data display from my database but other shows but not on a table
請仔細查看我的代碼並幫助我修復代碼中的錯誤。 只有我的第一個數據顯示在我的視圖頁面上的表格中,但其他數據顯示在視圖頁面上而不是表格上。
我的圖像名稱和圖像本身已成功插入我的數據庫和我的圖像目錄,我將其命名為“上傳”,但圖像不會顯示在我的視圖頁面上。
<?php
include ("config.php");
// Retrieve data from database
$sql="SELECT * FROM $tbl_name";
$result=mysql_query($sql);
echo "<table border='1'>
<tr>
<th>id</th>
<th>firstname</th>
<th>lastname</th>
<th>address</th>
<th>nationality</th>
<th>accountnumber</th>
<th>accounttype</th>
<th>balance</th>
<th>passport</th>
<th>username</th>
<th>passport</th>
<th>update</th>
<th>delete</th>
</tr>";
while($row = mysql_fetch_array($result)){
echo "<tr>";
echo "<td>" . $row['id'] . "</td>";
echo "<td>" . $row['firstname'] . "</td>";
echo "<td>" . $row['lastname'] . "</td>";
echo "<td>" . $row['address'] . "</td>";
echo "<td>" . $row['nationality'] . "</td>";
echo "<td>" . $row['account'] . "</td>";
echo "<td>" . $row['accounttype'] . "</td>";
echo "<td>" . $row['balance'] . "</td>";
echo "<td><h1><img src=\"upload/\" height=35 width=35 /> $row[id]</h1></td>";
echo "<td>" . $row['username'] . "</td>";
echo "<td>" . $row['password'] . "</td>";``
echo "<td><a href=\"update.php?id=" . $row['id'] . "\">update</a></td>";
echo "<td><a href=\"delete.php?id=" . $row['id'] . "\">delete</a></td>";
echo "</table>";
// close while loop
}
?>
通過快速查看,我看到你正在關閉while循環中的表。 你應該改變它
</tr>
並記住使用thead和tbody標簽。 結束循環后關閉tbody和table。
這需要在你的while循環之外。
echo "</table>";
和
<img src=\"upload/\"
僅指向上載目錄,您沒有指定實際圖像。 嘗試類似的東西:
echo "<td><h1><img src=\"upload/$row['image']\" height=35 width=35 /> $row['id']</h1></td>";
將表關閉選項卡移出while循環將解決第一個問題。
更改代碼如下以解決圖像問題。
echo '<td><img src="upload/'.$row['your image name'].'" height=35 width=35 ></td>';
您的代碼應如下所示。
<?php
include ("config.php");
// Retrieve data from database
$sql="SELECT * FROM $tbl_name";
$result=mysql_query($sql);
echo "<table border='1'>
<tr>
<th>id</th>
<th>firstname</th>
<th>lastname</th>
<th>address</th>
<th>nationality</th>
<th>accountnumber</th>
<th>accounttype</th>
<th>balance</th>
<th>passport</th>
<th>username</th>
<th>passport</th>
<th>update</th>
<th>delete</th>
</tr>";
while($row = mysql_fetch_array($result)){
echo "<tr>";
echo "<td>" . $row['id'] . "</td>";
echo "<td>" . $row['firstname'] . "</td>";
echo "<td>" . $row['lastname'] . "</td>";
echo "<td>" . $row['address'] . "</td>";
echo "<td>" . $row['nationality'] . "</td>";
echo "<td>" . $row['account'] . "</td>";
echo "<td>" . $row['accounttype'] . "</td>";
echo "<td>" . $row['balance'] . "</td>";
echo '<td><img src="upload/'.$row['your image name'].'" height=35 width=35 ></td>';
echo "<td>" . $row['username'] . "</td>";
echo "<td>" . $row['password'] . "</td>";``
echo "<td><a href=\"update.php?id=" . $row['id'] . "\">update</a></td>";
echo "<td><a href=\"delete.php?id=" . $row['id'] . "\">delete</a></td>";
echo "</tr>";
// close while loop
}
echo "</table>";
?>
我在更新頁面上顯示我的身份信息有點困難。 我有兩個條目6和7,如果我點擊ID 6上的更新,它將顯示在我的更新頁面上,但如果我點擊地址欄上的id 7更新,它將指示id = 7但它仍將顯示id 6條信息。
這是我的顯示代碼,在我的更新頁面上顯示數據
<?ph
include ('config.php');
// Retrieve data from database
$sql="SELECT * FROM $tbl_name ";
$result=mysql_query($sql);
$row=mysql_fetch_array($result);
?>
其次,如果我更改數據我的更新頁面,更改將不會反映在我的視圖頁面上。
<?php
include('config.php');
//This is the directory where images will be saved
$firstname=$_POST['firstname'];
$lastname=$_POST['lastname'];
$address=$_POST['address'];
$nationality=$_POST['nationality'];
$accountnumber=$_POST['account'];
$accounttype=$_POST['accounttype'];
$balance=$_POST['balance'];
$username=$_POST['username'];
$password=$_POST['password'];
$id=$_POST['id'];
// update data in mysql database
$sql="UPDATE $tbl_name SET firstname='$firstname', lastname='$lastname', address='$address', nationality='$nationality',accountnumber='$account',accounttype='$accounttype',balance='$balance',username='$username',password='$password'
WHERE id='$id'";
$result=mysql_query($sql);
header ("Location: details.php");
?>
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