用html表单和php更新mysql行 [英] update mysql row with html form and php

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问题描述

我一直在浏览这里的很多主题，但没有找到解决我的问题的方法。
我创建了一个表单，用于显示输入框中的数据库内容，当我更改内容时，它应该在数据库中更新。

 <？php 
 $ con = mysqli_connect（localhost ， 根， ， 头版）; 
 //检查连接
 if（mysqli_connect_errno（））{
 echo无法连接到MySQL：。 mysqli_connect_error（）; 
 
 
 $ result = mysqli_query（$ con，SELECT * FROM frontpage_left_links）
或die（Error：.mysqli_error（$ con））; 
 
 while（$ row = mysqli_fetch_array（$ result））{
 echo'< form action =method =post>'; 
 echo'< div style =float：left>'; 
 echo'< table border =1bordercolor =＃000000>'; 
 echo'< tr>'; 
 echo'< td> link< / td>'; 
 echo'< td>< input type =textname =linkidvalue ='。$ row ['link']。'>< / td>'; 
 echo'< / tr>'; 
 echo'< tr>'; 
 echo'< td> img< / td>'; 
 echo'< td>< input type =textname =imgidvalue ='。$ row ['img']。'>< / td>'; 
 echo'< / tr>'; 
 echo'< tr>'; 
 echo'< td> tekst< / td>'; 
 echo'< td>< input type =textname =imgidvalue ='。$ row ['name']。'>< / td>'; 
 echo'< / tr>'; 
 echo'< tr>'; 
 echo'< td>< input type =submitid =updatename =gemvalue =Gem< / td>< / td>'; 
 echo'< td>< input type =hiddenname =idvalue ='。$ row ['id']。'>< / td>'; 
 echo'< / tr>'; 
 echo'< / table>< / div>'; 
 echo'< div style =float：left>< a href ='。$ row ['link']。'>< center>< img src =img / '。$ row ['img']。'>< br />'.$row['name'].'</center></a></div>'; 
 echo'< / form>< br />< br />< br />< br>< br />< br />< br />'; 
} 
 
 if（isset（$ _ POST ['update']））{
 
 $ id = $ _POST ['id']; 
 $ link = $ _POST ['linkid']; 
 $ img = $ _POST ['imgid']; 
 $ name = $ _POST ['nameid']; 
 
 $ sql = mysqli_query（UPDATE frontpage_left_links SET link ='$ link'，img ='$ img'，name ='$ name'WHERE id ='$ id'）; 
 
 $ retval = mysqli_query（$ sql，$ con）; 
 
 if（！$ retval）{
 die（'Could not update data：'。mysql_error（））; 
} 
 
 echo已成功更新数据\; 
} 
 
 mysqli_close（$ con）; 
？>

表单显示数据库内容正常，但更改后没有任何反应。



感谢您的帮助。





这就是现在的样子。 

 <？php 
 $ con = mysqli_connect（localhost，root，，frontpage ）; 
 //检查连接
 if（mysqli_connect_errno（））
 {
 echo无法连接到MySQL：。 mysqli_connect_error（）; 
} 
 
 if（isset（$ _ POST ['gem']））
 {
 
 $ id = $ _POST ['id']; 
 $ link = $ _POST ['linkid']; 
 $ img = $ _POST ['imgid']; 
 $ name = $ _POST ['nameid']; 
 
 $ sql = mysqli_query（UPDATE frontpage_left_links SET link ='$ link'，img ='$ img'，name ='$ name'WHERE id ='$ id'）; 
 
 $ retval = mysqli_query（$ con，$ sql）; 
 if（！$ retval）
 {
 die（'Could not update data：'。mysql_error（））; 
} 
 echo已成功更新数据\; 
 
 
 
 $ b $ result = mysqli_query（$ con，SELECT * FROM frontpage_left_links）
或者die（Error：.mysqli_error（ $ CON））; 
 
 while（$ row = mysqli_fetch_array（$ result））
 {
 echo'< form action =method =post>'; 
 echo'< div style =float：left>'; 
 echo'< table border =1bordercolor =＃000000>'; 
 echo'< tr>'; 
 echo'< td> link< / td>'; 
 echo'< td>< input type =textname =linkidvalue ='。$ row ['link']。'>< / td>'; 
 echo'< / tr>'; 
 echo'< tr>'; 
 echo'< td> img< / td>'; 
 echo'< td>< input type =textname =imgidvalue ='。$ row ['img']。'>< / td>'; 
 echo'< / tr>'; 
 echo'< tr>'; 
 echo'< td> tekst< / td>'; 
 echo'< td>< input type =textname =nameidvalue ='。$ row ['name']。'>< / td>'; 
 echo'< / tr>'; 
 echo'< tr>'; 
 echo'< td>< input type =submitid =updatename =gemvalue =Gem< / td>< / td>'; 
 echo'< td>< input type =hiddenname =idvalue ='。$ row ['id']。'>< / td>'; 
 echo'< / tr>'; 
 echo'< / table>< / div>'; 
 echo'< div style =float：left>< a href ='。$ row ['link']。'>< center>< img src =img / '。$ row ['img']。'>< br />'.$row['name'].'</center></a></div>'; 
 echo'< / form>< br />< br />< br />< br>< br />< br />< br />'; 
} 
 
 
 
 mysqli_close（$ con）; 
？>

现在我收到这个错误消息。

警告：mysqli_query（）需要至少2个参数，1在第17行的/Applications/XAMPP/xamppfiles/htdocs/page/admin.php中给出



警告：mysqli_query ）：在第19行中的/Applications/XAMPP/xamppfiles/htdocs/page/admin.php中的空查询
无法更新数据：

解决方案

因为你的更新是在页面的最后，所以把它放在其余的上面。



还要更改isset（$ _ POST ['update'] isset（$ _ POST ['gem'] 

 <？php 
 $ con = mysqli_connect（localhost， root，，frontpage）; 
 //检查连接
 if（mysqli_connect_errno（））
 {
 echo无法连接到MySQL：mysqli_connect_error （）; 
} 
 
 if（isset（$ _ POST ['gem']））
 {
 
 $ id = $ _POST ['id ']; 
 $ link = $ _POST ['linkid']; 
 $ img = $ _POST ['imgid']; 
 $ name = $ _POST ['nameid']; 
 
 $ sql =UPDATE frontpage_left_links SET link ='$ link'，img ='$ img'，name ='$ name'WHERE id ='$ id'; 
 
 $ retval = mysqli_query（$ con，$ sql）; 
 if（！$ retval）
 {
 die（'Could not update data：'。mysql_error（））; 
} 
 echo已成功更新数据\; 
 
 
 
 $ b $ result = mysqli_query（$ con，SELECT * FROM frontpage_left_links）
或者die（Error：.mysqli_error（ $ CON））; 
 
 while（$ row = mysqli_fetch_array（$ result））
 {
 echo'< form action =method =post>'; 
 echo'< div style =float：left>'; 
 echo'< table border =1bordercolor =＃000000>'; 
 echo'< tr>'; 
 echo'< td> link< / td>'; 
 echo'< td>< input type =textname =linkidvalue ='。$ row ['link']。'>< / td>'; 
 echo'< / tr>'; 
 echo'< tr>'; 
 echo'< td> img< / td>'; 
 echo'< td>< input type =textname =imgidvalue ='。$ row ['img']。'>< / td>'; 
 echo'< / tr>'; 
 echo'< tr>'; 
 echo'< td> tekst< / td>'; 
 echo'< td>< input type =textname =imgidvalue ='。$ row ['name']。'>< / td>'; 
 echo'< / tr>'; 
 echo'< tr>'; 
 echo'< td>< input type =submitid =updatename =gemvalue =Gem< / td>< / td>'; 
 echo'< td>< input type =hiddenname =idvalue ='。$ row ['id']。'>< / td>'; 
 echo'< / tr>'; 
 echo'< / table>< / div>'; 
 echo'< div style =float：left>< a href ='。$ row ['link']。'>< center>< img src =img / '。$ row ['img']。'>< br />'.$row['name'].'</center></a></div>'; 
 echo'< / form>< br />< br />< br />< br>< br />< br />< br />'; 
} 
 
 
 
 mysqli_close（$ con）; 
？>


I've been looking through many threads on here without finding a solution to my problem. I've created a form that is supposed to show content of a database in input boxes, and when i change the content, it should be updated in the database.

No errors, nothing gets changed.

<?php
$con=mysqli_connect("localhost","root","","frontpage");
// Check connection
if (mysqli_connect_errno()){
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$result = mysqli_query($con,"SELECT * FROM frontpage_left_links")
or die("Error: ".mysqli_error($con));

while($row = mysqli_fetch_array($result)){
    echo '<form action="" method="post">';
    echo '<div style="float:left">';
    echo '<table border="1" bordercolor="#000000">';
    echo '<tr>';
    echo '<td>link</td>';
    echo '<td><input type="text" name="linkid" value="'.$row['link'].'"></td>';
    echo '</tr>';
    echo '<tr>';
    echo '<td>img</td>';
    echo '<td><input type="text" name="imgid" value="'.$row['img'].'"></td>';
    echo '</tr>';
    echo '<tr>';
    echo '<td>tekst</td>';
    echo '<td><input type="text" name="imgid" value="'.$row['name'].'"></td>';
    echo '</tr>';
    echo '<tr>';
    echo '<td><input type="submit" id="update" name="gem" value="Gem"</td></td>';
    echo '<td><input type="hidden" name="id" value="'.$row['id'].'"></td>';
    echo '</tr>';
    echo '</table></div>';
    echo '<div style="float:left"><a href="'.$row['link'].'"><center><img src="img/'.$row['img'].'"><br />'.$row['name'].'</center></a></div>';
    echo '</form><br /><br /><br /><br /><br /><br /><br /><br />';
}

if(isset($_POST['update'])){

    $id = $_POST['id'];
    $link = $_POST['linkid'];
    $img = $_POST['imgid'];
    $name = $_POST['nameid'];

    $sql = mysqli_query("UPDATE frontpage_left_links SET link = '$link', img = '$img', name = '$name' WHERE id = '$id'");

    $retval = mysqli_query( $sql, $con );

    if(! $retval ){
        die('Could not update data: ' . mysql_error());
    }

    echo "Updated data successfully\n";
}

mysqli_close($con);
?>

The form show the database content fine, but nothing happens when changed.

I appreciate any help I can get.

This is what it looks like now.

<?php
$con=mysqli_connect("localhost","root","","frontpage");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

if(isset($_POST['gem']))
{

$id = $_POST['id'];
$link = $_POST['linkid'];
$img = $_POST['imgid'];
$name = $_POST['nameid'];

$sql = mysqli_query("UPDATE frontpage_left_links SET link = '$link', img = '$img', name = '$name' WHERE id = '$id'");

$retval = mysqli_query( $con, $sql );
if(! $retval )
{
  die('Could not update data: ' . mysql_error());
}
echo "Updated data successfully\n";


}

$result = mysqli_query($con,"SELECT * FROM frontpage_left_links")
or die("Error: ".mysqli_error($con));

while($row = mysqli_fetch_array($result))
  {
  echo '<form action="" method="post">';
  echo '<div style="float:left">';
  echo '<table border="1" bordercolor="#000000">';
  echo '<tr>';
  echo '<td>link</td>';
  echo '<td><input type="text" name="linkid" value="'.$row['link'].'"></td>';
  echo '</tr>';
  echo '<tr>';
  echo '<td>img</td>';
  echo '<td><input type="text" name="imgid" value="'.$row['img'].'"></td>';
  echo '</tr>';
  echo '<tr>';
  echo '<td>tekst</td>';
  echo '<td><input type="text" name="nameid" value="'.$row['name'].'"></td>';
  echo '</tr>';
  echo '<tr>';
  echo '<td><input type="submit" id="update" name="gem" value="Gem"</td></td>';
  echo '<td><input type="hidden" name="id" value="'.$row['id'].'"></td>';
  echo '</tr>';
  echo '</table></div>';
  echo '<div style="float:left"><a href="'.$row['link'].'"><center><img src="img/'.$row['img'].'"><br />'.$row['name'].'</center></a></div>';
  echo '</form><br /><br /><br /><br /><br /><br /><br /><br />';
  }



mysqli_close($con);
?>

Now i get this error.

Warning: mysqli_query() expects at least 2 parameters, 1 given in /Applications/XAMPP/xamppfiles/htdocs/page/admin.php on line 17

Warning: mysqli_query(): Empty query in /Applications/XAMPP/xamppfiles/htdocs/page/admin.php on line 19 Could not update data:

解决方案

Because your udate is at the end of the page put it above the rest.

And also change isset($_POST['update'] to isset($_POST['gem']

<?php
$con=mysqli_connect("localhost","root","","frontpage");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

if(isset($_POST['gem']))
{

$id = $_POST['id'];
$link = $_POST['linkid'];
$img = $_POST['imgid'];
$name = $_POST['nameid'];

$sql = "UPDATE frontpage_left_links SET link = '$link', img = '$img', name = '$name' WHERE id = '$id'";

$retval = mysqli_query($con,$sql );
if(! $retval )
{
  die('Could not update data: ' . mysql_error());
}
echo "Updated data successfully\n";


}

$result = mysqli_query($con,"SELECT * FROM frontpage_left_links")
or die("Error: ".mysqli_error($con));

while($row = mysqli_fetch_array($result))
  {
  echo '<form action="" method="post">';
  echo '<div style="float:left">';
  echo '<table border="1" bordercolor="#000000">';
  echo '<tr>';
  echo '<td>link</td>';
  echo '<td><input type="text" name="linkid" value="'.$row['link'].'"></td>';
  echo '</tr>';
  echo '<tr>';
  echo '<td>img</td>';
  echo '<td><input type="text" name="imgid" value="'.$row['img'].'"></td>';
  echo '</tr>';
  echo '<tr>';
  echo '<td>tekst</td>';
  echo '<td><input type="text" name="imgid" value="'.$row['name'].'"></td>';
  echo '</tr>';
  echo '<tr>';
  echo '<td><input type="submit" id="update" name="gem" value="Gem"</td></td>';
  echo '<td><input type="hidden" name="id" value="'.$row['id'].'"></td>';
  echo '</tr>';
  echo '</table></div>';
  echo '<div style="float:left"><a href="'.$row['link'].'"><center><img src="img/'.$row['img'].'"><br />'.$row['name'].'</center></a></div>';
  echo '</form><br /><br /><br /><br /><br /><br /><br /><br />';
  }



mysqli_close($con);
?>

这篇关于用html表单和php更新mysql行的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！

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