如何将id发布到另一个页面 [英] how to post id to another page

问题描述

嗨
我是新手，请帮我解决以下问题。单击更新按钮时，ID应该是另一个页面来处理更新功能。以下是我的全部代码



index.php

 <？php  
 include_once '  ../。 ./inc/config.inc.php'; 
 
 $ sql =   SELECT * FROM demo order by id; 
 尝试 {
  $ result  = mysqli_query（$ con，$ sql）; 
包括'  view.html.php'; 
} 
  catch （PDOException $ e）
 {
  echo  $ e-> getMessage（）。   \ n; 
 file_put_contents（'  PDOErrors.txt'，$ e-> getMessage（），FILE_APPEND ）; 
  //   $ output ='从数据库中提取作者时出错！';  
  //  包含'../../ notification / errormsg.php';  
退出（）; 
} 

view.html.php



< pre lang =PHP> < 脚本 语言 = javascript < span class =code-keyword>>
function change_action（）
{
var frm_obj = document .getElementById（ FRM）;
frm_obj.action = data.php;
}
< / 脚本 >

< h3>用户详细信息< / h3 >
< form action = method = POST id = frm>
< table width = 100％ align = center cellpadding = 4 cellspacing = 1>
< tr>

< td> ID < / td >
< td> ID < / td >
< td> NAME < / td >
< td> FIRST NAME < / td >
< td> AGE < / td >
< td> < / td >

< / tr >
<？php
if （isset（$ result））{
while （$ row = mysqli_fetch_array（$ result））{？>
< tr>

< td>< input type = text name = vid value = <？php echo $ row ['id'];？> /> < / < span class =code-leadattribute> td >
< td><？php echo $ row [ id'];？> < / < span class =code-leadattribute> td >
< td><？php echo $ row [ name'];？> < / < span class =code-leadattribute> td >
< td><？php echo $ row [' firstname'];？> < / td >
< td><？php echo $ row [' age'];？> < / td >
< td>< input type = submit value = update name = update onclick = change_action（）>
< input type = submit value = delete name = 删除 onclick = change_action（）>
< / td >
< / tr >
<？php
}

}
mysqli_close（$ con）;
？>

< / table >
< / FORM >
<？php
包括' ../。 ./inc/footer.php';

data.php

 <？php  
 include_once '  ../../ inc / config.inc.php'; 
  if （isset（$ _ POST [' 更新']）&& $ _POST ['  update']！= < span class =code-string> ）
 {
  $ id  = $ _POST ['  vid']; 
 $ sql =   SELECT * FROM demo where id ='$ id'; 
  $ result  = mysqli_query（$ con，$ sql）; 
包括'  edit.php'; 
} 

edit.php

 <？php  
  echo  ' < span class =code-string> edit.php'; 
 
 if（isset（$ result））
 {
 while（$ row = mysqli_fetch_array（$ result））
 {
  echo  $ row ['  id'];  < br>'; 
  echo  $ row ['  name']; ' < br>'; 
  echo  $ row ['  firstname']; ' < br>'; 
  echo  $ row ['  age']; ' < br>'; 
 
} 
} 

谢谢



maideen

解决方案

sql = 通过id; SELECT * FROM演示订单;
尝试 {

结果 = mysqli_query（

con，

Hi I am new, Pls help me the following issue. When click the update button, ID should be to another page to process the update function. Below is my entire code

index.php

<?php
include_once '../../inc/config.inc.php';

$sql="SELECT * FROM demo  order by id";
    try {
            $result = mysqli_query($con,$sql);  
            include 'view.html.php'; 
         } 
    catch (PDOException $e) 
        {
           echo $e->getMessage() . "\n";
           file_put_contents('PDOErrors.txt', $e->getMessage(), FILE_APPEND);
//           $output = 'Error fetching authors from database!';
//           include '../../notification/errormsg.php';
           exit();
        }

view.html.php

<Script Language="javascript">
function change_action()
    {
        var frm_obj=document.getElementById("frm");
        frm_obj.action="data.php";
    }
</Script>

<h3>User Details</h3>
<form action="" method="POST" id="frm" >
    <table width="100%" align="center" cellpadding="4" cellspacing="1">
        <tr>
            
            <td>ID</td>
            <td>ID</td>
            <td>NAME</td>
            <td>FIRST NAME</td>
            <td>AGE</td>
            <td></td>
            
        </tr>
        <?php
            if(isset($result)){
            while($row = mysqli_fetch_array($result)){ ?>
             <tr>
                
                <td><input type="text" name="vid" value="<?php echo $row['id'];?>"/></td>
                <td><?php echo $row['id'];?></td>
                <td><?php echo $row['name'];?></td>
                <td><?php echo $row['firstname'];?></td>
                <td><?php echo $row['age'] ;?></td>
                <td><input type="submit" value="update" name="update" onclick="change_action()">
                <input type="submit" value="delete" name="delete" onclick="change_action()">
                </td>                
              </tr>  
         <?php
            } 
         
            }
            mysqli_close($con);
        ?>
   
    </table>
</FORM>
<?php
include '../../inc/footer.php';

data.php

<?php
include_once '../../inc/config.inc.php';
if (isset($_POST['update']) && $_POST['update']  != "" )
        {
            $id = $_POST['vid'];
            $sql="SELECT * FROM demo where id='$id'";
            $result = mysqli_query($con,$sql);
            include 'edit.php';
     }

edit.php

<?php
echo 'edit.php';
  
if(isset($result))
    {
        while($row = mysqli_fetch_array($result))
                { 
                    echo $row['id'];'<br>';
                    echo $row['name'];'<br>';
                    echo $row['firstname'];'<br>';
                    echo $row['age'];'<br>';
    
                }
    }

thank you

maideen

解决方案

sql="SELECT * FROM demo order by id"; try {

result = mysqli_query(

con,

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