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更新fork Github(Update the fork Github)

 我已经分叉了一个远程回购，它有3个分支 
 

     
 
  1.  主 
    2.  
 
  2.  一个 
    3.  
 
  3.  b 
    4.  

 

 但是该团队向遥控器添加了新的分支。 目前的分支机构是 
 

     
 
  1.  主 
    2.  
 
  2.  一个 
    3.  
 
  3.  b 
    4.  
 
  4.  C 
    5.  
 
  5.  d 
    6.  
 
  6.  Ë 
    7.  

 

 但我的叉子只有3个分支。 我应该如何更新我的fork以包含远程仓库中存在的所有分支（包括新增加的分支）。 

I have forked a remote repo and it has 3 branches 
 

     
 
  1. master
    2.  
 
  2. a
    3.  
 
  3. b
    4.  

 

But the team added new branches to the remote. And the current branches are
 

     
 
  1. master
    2.  
 
  2. a
    3.  
 
  3. b
    4.  
 
  4. c
    5.  
 
  5. d
    6.  
 
  6. e
    7.  

 

but my fork has only 3 branches. How should I update my fork to contain all the branches (including newly added branches) present in remote repo.

原文：https://stackoverflow.com/questions/15738604
更新时间：2023-12-04 15:12

最满意答案

 尝试这个 
 

<?php
    session_start();
    function putForm2(){
    $myForm = "<form name='form2' method='post' action='#'><table border='1'><tr><td>form 2 is here!<br></td></tr><tr><td><input type='text' name='txt123' id='txt123'></td></tr> <tr><td><input type='submit' name='sendtwo' id='sendtwo' value='TwoClick'></td></tr></table></form>";
    return $myForm;
    }
?><!DOCTYPE html>
    <html>
<head><title></title>
</head>
<body>
<div id="One">
    <form name="form1" method="post" action="#">
        <?php echo "<input type='text' name='txt1' id='txt1'>"; // This text box is dynamically populated ?>
              <input type="submit" name="sendone" id="sendone" value="OneClick">
    </form>
</div>

<div id="two">


            <?php
            if(isset($_POST['sendone']))
            {   if($_POST['txt1'] == '')
                {echo 'txt1 is empty!'; return;} else {$_SESSION['txt1'] = $_POST['txt1'];}

                        if(isset($_SESSION['txt1'])) 
                        echo $_SESSION['txt1']; 
                        echo putForm2();



            }   
                        if(isset($_POST['sendtwo']))
                        if(isset($_POST['sendtwo'])) 
                        {
                            if($_POST['txt123'] == '')
                            {
                                echo putForm2();
                                echo "Text box is empty..."; return;
                            }
                        }               
                        ?>

        </div>
    </body>
</html>


try this
 

<?php
    session_start();
    function putForm2(){
    $myForm = "<form name='form2' method='post' action='#'><table border='1'><tr><td>form 2 is here!<br></td></tr><tr><td><input type='text' name='txt123' id='txt123'></td></tr> <tr><td><input type='submit' name='sendtwo' id='sendtwo' value='TwoClick'></td></tr></table></form>";
    return $myForm;
    }
?><!DOCTYPE html>
    <html>
<head><title></title>
</head>
<body>
<div id="One">
    <form name="form1" method="post" action="#">
        <?php echo "<input type='text' name='txt1' id='txt1'>"; // This text box is dynamically populated ?>
              <input type="submit" name="sendone" id="sendone" value="OneClick">
    </form>
</div>

<div id="two">


            <?php
            if(isset($_POST['sendone']))
            {   if($_POST['txt1'] == '')
                {echo 'txt1 is empty!'; return;} else {$_SESSION['txt1'] = $_POST['txt1'];}

                        if(isset($_SESSION['txt1'])) 
                        echo $_SESSION['txt1']; 
                        echo putForm2();



            }   
                        if(isset($_POST['sendtwo']))
                        if(isset($_POST['sendtwo'])) 
                        {
                            if($_POST['txt123'] == '')
                            {
                                echo putForm2();
                                echo "Text box is empty..."; return;
                            }
                        }               
                        ?>

        </div>
    </body>
</html>

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