提交按钮提交错误的表单 [英] Submit button submitting wrong form

问题描述

我为每个表单提供了2个表单和2个提交按钮。现在问题是每当我点击第二个表格的第二个提交按钮时，该提交按钮正在提交第一个表格！两个提交按钮提交第一个表格！我不知道该怎么办

ps - 第二个提交按钮仅在只有一个表格时有效

 <？php  include（'  D：\websites\rushi\html \include\adminmain.php'）; ？>  
 
 .width_3_quarter1 {
 width：100 %%; 
 margin-right：100px; 
 float：left; 
} 
 <   h3     class   =  tabs_involved    style   =  margin-top：10px > 学生列表<   / h3  >  
 
  <？ php 
  $ servername  = ' 本地主机'; 
  $ username  = '  root'; 
  $ password  = ' '; 
  $ conn  = mysql_connect（$ servername，$ username，$ password）; 
 
 if（$ conn === false）{
 die（  ERROR ：无法连接。 .mysqli_connect_error（））; 
} 
 
 mysql_select_db（'  school'）; 
 
  $ query  =   select * from class; 
 
  $ result  = mysql_query（$ query）; 
 
  echo   选择班级：; 
  echo   ; 
  echo   ; 
 
 while（$ row = mysql_fetch_array（$ result））
 {
  echo   。$ row ['  class_name']。 ; 
} 
  echo    ; 
 
  echo   < br> 
< input type ='submit'
 value ='search'
 class = \alt_btn \
 name ='search'
> ;; 
  echo   < br>< BR>中; 
  ; 
 
 ？>  
 
 
 <？  php 
 
  $ servername  = ' 本地主机'; 
  $ username  = '  root'; 
  $ password  = ' '; 
 
  $ conn  = mysql_connect（$ servername，$ username，$ password）; 
 
 $ _SESSION ['  id'] = $ id; 
 
  $ search  = $ _POST ['  SRCH']; 
 
  $ sql  =   SELECT c。*，a。* FROM class c，tedence a WHERE c.student_id = a.student_id AND class_name ='$ search'; 
 
 mysql_select_db（'  school'）; 
 
  $ ds  = mysql_query（$ sql，$ conn）; 
 
  echo   < form method = \post \name ='myForm'id ='myForm'action ='cuo.php'>; 
 
  $ mn  = $ ds ['  ID ]; 
 
  if （isset（$ _POST ['  srch']））{
 while（$ dr = mysql_fetch_array（$ ds））
 {
  print   ; 
  print   ; 
  print   ; 
  print   ; 
  print   ; 
  print   ; 
  print   ; 
 
  echo  isset（$ _ POST ['  id']）？ '  checked =checked'：' '; 
  print   ; 
} 
} 
  print   < table class = tablesorter   cellspacing =  0   >< thead>< tr>< th> ;类标识码< /第><的第i; CLASS_NAME< /第><的第i; student_name< /第><的第i; student_id数据< /第><的第i; TEACHER_NAME< /第><的第i;状态< /次> ;< th> date< / th>< / tr>< / thead>< tbody>< tr>< td width =  100   >。（$ dr [   class_id]）。 < ; / td>< td width =  100   >。（$ dr [  class_name]）。 < / td>< td width =  100   >。（$ dr [  student_name]）。 < / td>< td width =  100   >。（$ dr [  student_id]). < / td>< td width =  100   >。（$ dr [  teacher_name]）。 < / td>< td width =  100   >< input type = \checkbox \name = \attd [] \ value =。$ dr ['  id']。 < / td>< td width =  100  < span class =code-string> >。（$ dr [  date]）。 < / td> < / TR GT&;< / tbody的>< /表>; 
  echo   < br> 
< input type ='submit'
 value ='update'
 class = \alt_btn \
 name ='update'
>; 
  echo   ; 
 
 mysql_close（$ conn）; 
 ？>  

我有什么试过：



我试过javascript，jquery，ajax但没什么用呢！

解决方案

< blockquote> servername = ' localhost';

用户名 = ' 根;

密码 = ' ';

I have 2 forms and 2 submit button for each form. now the problem is whenever i click on 2nd submit button of 2nd form,that submit button is submiting 1st form! both submit button submit 1st form! i dont know what to do now
ps - 2nd submit button works only when there is only one form

<?php include('D:\websites\rushi\html\include\adminmain.php'); ?>

.width_3_quarter1 {
width: 100%%;
margin-right: 100px;
float: left;
}
		<h3 class="tabs_involved" style="margin-top: 10px">student list</h3>                    
			 
<?php 
$servername = 'localhost';
$username = 'root';
$password = '';
$conn = mysql_connect($servername, $username, $password);
 
if($conn === false){
    die("ERROR: Could not connect. " . mysqli_connect_error());
}
 
 mysql_select_db('school');

$query = "select * from class";

$result = mysql_query($query);	

echo "Select class : ";
echo "";
echo "";

 while($row=mysql_fetch_array($result))
 { 
       echo "".$row['class_name']."";
 }
 echo "";

echo "<br>
      <input type='submit'
	  value = 'search'
	  class=\"alt_btn\"
      name='search' 
      >";
 echo "<br><br>";
"";

 ?>


<?php

$servername = 'localhost';
$username = 'root';
$password = '';

$conn = mysql_connect($servername, $username, $password);
    
		 $_SESSION['id']=$id; 

$search = $_POST['srch'];
	
$sql = "SELECT c.*,a.* FROM class c, attendence a WHERE c.student_id=a.student_id AND class_name = '$search'";

mysql_select_db('school');

$ds = mysql_query($sql,$conn);

echo "<form  method = \"post\" name ='myForm' id ='myForm' action='cuo.php'>";

$mn = $ds['id'];

if (isset ($_POST['srch'])){
while($dr = mysql_fetch_array($ds))
{
	print"";
	print "";
	print ""; 
	print "";
	print "";
	print "";
	print "";
	
	 echo isset($_POST['id']) ? 'checked="checked" ' : '';
	print "";
	}	
}
  print "<table class="tablesorter" cellspacing="0"><thead> 	<tr><th>class_id</th><th>class_name</th><th>student_name</th><th>student_id</th><th>teacher_name</th><th>status</th><th>date</th></tr></thead><tbody><tr><td width="100">".($dr ["class_id"])."</td><td width="100">".($dr ["class_name"])."</td><td width="100">".($dr ["student_name"])."</td><td width="100">".($dr ["student_id"])."</td><td width="100">".($dr ["teacher_name"])."</td><td width="100"><input type=\"checkbox\" name=\"attd[]\" value=".$dr['id']."</td><td width="100">".($dr ["date"])."</td></tr></tbody></table>";
echo "<br>
      <input type='submit'
	  value = 'update'
	  class=\"alt_btn\"
	  name='update'
      >";
  echo "";

mysql_close($conn);
?>

What I have tried:

i've tried javascript,jquery,ajax but nothing works!

解决方案

servername = 'localhost';

username = 'root';

password = '';

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