php表单提交 - Q2 [英] php form submit - Q2

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问题描述

对于虚拟问题,我很抱歉。这里是我简单的PHP表单,带有两个SQL表格和ADD submit按钮,我想将人员从Test1移动到Test2。许多事情都很好:( 只有提交按钮不起作用,因此没有来自Test2表的反馈。



修正:现在提交效果很好



Q2 - 仍然无法使用复选框:( - 请



有人可以告诉我如何追踪这样的错误吗?

 < php include(db_connect.php);?> 
< html>
< head>< / head>
< body>
< form method =postaction =moving.php>
< table border =1>
< tr>
< td>
< php
$ left ='SELECT * FROM test1 ORDER BY name ASC';
$ result1 = mysql_query($ left);
$ count = mysql_num_rows($ result1);
while ($ resulta = mysql_fetch_array($ result1)){
?>
< input name =checkbox []type =checkboxid = 复选框[]value =<? echo $ resulta ['id']; >>< ;? echo $ resulta ['name'];?>
<>
<?php}>
< / td>
< td>< input type =submitid =addname =addvalue =Add/>< / td>
< td>
<?php
$ rigth ='SELECT * FROM test2,test1 WHERE test2.collect = test1.id ORDER BY test1.name ASC';
$ result2 = mysql_query($ right) ;
while($ resultb = mysql_fetch_array($ result2)){
echo $ resultb ['id'];
echo< br />;
}
?>
< / td>
< / tr>
< / table>
<?php
//检查add ($ i = 0; $ i <$ count; $ i ++){
$ add_id){is $($ _ POST ['add'])){
启动此
= $ checkbox [$ i];
if($ add_id =='1'){
$ sql =INSERT INTO test2(status,collect)VALUES(1,1);
$ result = mysql_query($ sql);



//如果成功重定向到delete_multiple.php
if($ result){
echo< meta http-equiv = \refresh \content = \0; URL = moving.php\>;
}
}
mysql_close();
?>
< / form>
< / body>
< / html>

感谢:)初学者

解决方案

解决方案 - 不是很好,但它的工作原理 - 感谢所有的意见和帮助!!!

 <?php include(db_connect.php);?> 

< html>
< head>
< / head>
< body>
< form method =postaction =test.php>

新:

 <?php 
$ left ='SELECT * FROM test1 ORDER BY name ASC';
$ result1 = mysql_query($ left);
$ count = mysql_num_rows($ result1);
while($ resulta = mysql_fetch_array($ result1))
{
?>

< input name =checkbox_add []type =checkboxid =checkbox_add []value =< ;? echo $ resulta ['id'];?>/ > < ;? echo $ resulta ['name']; ?>
< br />
<?php
}
?>

< / td> < td>< input type =submitid =addname =addvalue =Add/>< br />< input type =submitid =deletename =删除值=Del/>< / td>< td>
<?php
$ right ='SELECT test2.id,test1.name FROM test2,test1 WHERE test1.id = test2.collect AND test2.status = 1';
$ result2 = mysql_query($ right);
while($ resultb = mysql_fetch_array($ result2))
{
?>
< input name =checkbox_del []type =checkboxid =checkbox_del []value =<?php echo $ resultb ['id'];?>/>, <?php echo $ resultb ['id']; ?>,< ;? echo $ resultb ['name']; ?>
< br />
<?php
}
?>
< / td>< / tr>< / table>


<?php
//检查添加按钮是否有效,启动
if(isset($ _ POST ['add'])){
for($ c = 0; $ c< count($ _ POST ['checkbox_add']); $ c ++){
$ checkbox_add = $ _POST ['checkbox_add'] [$ c];
$ sql =INSERT INTO test2(status,collect)VALUES(1,。$ checkbox_add。);
echo $ sql;
$ result = mysql_query($ sql);
if($ result){
echo< meta http-equiv = \refresh'\\content = \0; URL = test.php\>;


$ b elseif(isset($ _ POST ['delete'])){
for($ c = 0; $ c $ checkbox_del = $ _POST ['checkbox_del'] [$ c];
回声日期(Y-m-d);
$ sql =UPDATE test2 SET status ='2',log ='。date('Y-m-d')。'其中id =。$ checkbox_del;
echo $ sql;
$ result = mysql_query($ sql);
if($ result){
echo< meta http-equiv = \refresh'\\content = \0; URL = test.php\>;


$ b elseif isset($ _ POST ['new'])){
$ sql =INSERT INTO test1(status,name)VALUES (1,'。$ _ POST ['newitem']。');
echo $ sql;
$ result = mysql_query($ sql);
if($ result){
echo< meta http-equiv = \refresh'\\content = \0; URL = test.php\>;
}
}
mysql_close();
?>

< / form>

< / body>
< / html>


i am sorry for the dummy question. Here is my simple PHP form with two SQL tables and with the ADD submit button I would like to move people from Test1 to Test2. Many thing are fine:( only the submit button does't work therefore no feedback from Test2 table.

Revised: Submit now works great

Q2 - still don't get the checkboxs to work:( - please

Could somebody show me how to track back such an error like this please?

<?php include("db_connect.php");?>
<html>
  <head></head>
  <body>
    <form method="post" action="moving.php">
      <table border="1">
        <tr>
          <td>
            <?php
              $left='SELECT * FROM test1 ORDER BY name ASC';
              $result1=mysql_query($left);
              $count=mysql_num_rows($result1);
              while($resulta = mysql_fetch_array($result1)) {
                  ?>
                  <input name="checkbox[]" type="checkbox" id="checkbox[]" value="<? echo $resulta['id']; ?>"/> <? echo $resulta['name']; ?>
                  <br />
            <?php } ?>
          </td>
          <td><input type="submit" id="add" name="add" value="Add" /></td>
          <td>
            <?php
              $rigth='SELECT * FROM test2,test1 WHERE test2.collect=test1.id ORDER BY test1.name ASC';
              $result2=mysql_query($right);
              while($resultb = mysql_fetch_array($result2)) {
                  echo $resultb['id'] ;
                  echo "<br />";
              }
            ?>
          </td>
        </tr>
      </table>
      <?php
        // Check if add button active, start this
        if (isset($_POST['add'])) {
            for ($i=0;$i<$count;$i++) {
                $add_id = $checkbox[$i];
                if ($add_id=='1') {
                  $sql = "INSERT INTO test2 (status, collect) VALUES(1, 1)";
                  $result = mysql_query($sql);
                }
            }

            // if successful redirect to delete_multiple.php
            if ($result){
                echo "<meta http-equiv=\"refresh\" content=\"0;URL=moving.php\">";
            }
        }
        mysql_close();
      ?>
    </form>
  </body>
</html>

thanks:) from a beginner

解决方案

the solution - not nice but it works - thanks for all the comments and help!!!

<?php include("db_connect.php");?>

<html>
<head>
</head>
<body>
<form method="post" action="test.php">

New:

<?php
$left='SELECT * FROM test1 ORDER BY name ASC';
$result1=mysql_query($left);
$count=mysql_num_rows($result1);
while($resulta = mysql_fetch_array($result1))
  {
?>

<input name="checkbox_add[]" type="checkbox" id="checkbox_add[]" value="<? echo $resulta['id']; ?>"/> <? echo $resulta['name']; ?>
<br />
<?php
   }
?>

</td> <td><input type="submit" id="add" name="add" value="Add" /><br /><input type="submit" id="delete" name="delete" value="Del" /></td><td>
<?php
$right='SELECT test2.id, test1.name FROM test2, test1 WHERE test1.id=test2.collect AND test2.status=1';
$result2=mysql_query($right);
while($resultb = mysql_fetch_array($result2))
  {
?>
   <input name="checkbox_del[]" type="checkbox" id="checkbox_del[]" value="<?php echo $resultb['id']; ?>"/>, <?php echo $resultb['id']; ?>, <? echo $resultb['name']; ?>
<br />
<?php
  }
?>
</td></tr></table>


<?php
// Check if add button active, start this
if (isset($_POST['add'])) {
  for ($c = 0; $c < count($_POST['checkbox_add']); $c++){
    $checkbox_add = $_POST['checkbox_add'][$c];
    $sql = "INSERT INTO test2 (status, collect) VALUES(1, ".$checkbox_add.")";
    echo $sql;
    $result = mysql_query($sql);
    if($result){
      echo "<meta http-equiv=\"refresh\" content=\"0;URL=test.php\">";
      }
    }
  }
elseif (isset($_POST['delete'])) {
  for ($c = 0; $c < count($_POST['checkbox_del']); $c++){
    $checkbox_del = $_POST['checkbox_del'][$c];
    echo date("Y-m-d");
    $sql = "UPDATE  test2 SET status='2', log='".date('Y-m-d')."' Where id=".$checkbox_del;
    echo $sql;
    $result = mysql_query($sql);
    if($result){
      echo "<meta http-equiv=\"refresh\" content=\"0;URL=test.php\">";
     }
    }
  }
elseif (isset($_POST['new'])) {
    $sql = "INSERT INTO test1 (status, name) VALUES(1, '".$_POST['newitem']."')";
    echo $sql;
    $result = mysql_query($sql);
    if($result){
      echo "<meta http-equiv=\"refresh\" content=\"0;URL=test.php\">";
    }
  }
mysql_close();
?>

</form>

</body>
</html>

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