为每一行打印提交按钮

Question

我的按钮有问题。我会尽力解释。 1) 我有来自 postgresql 的用户数据库和 ms sql 中的新数据库。 2) 在 table ("SELECT * from users"-postgresql ) 中创建了 2 列的站点：它们是 id/user 3) 然后添加了包含提交按钮的新列 "Operator"，提交按钮的功能正在从 ms sql 数据库更新 "access" 列。

问题：它打印我在 ms sql 中拥有的所有数据的所有按钮（我在 ms sql 中有 7 行数据，它为每行打印 7 个按钮），我需要 "echo" 1 个按钮对于每一行都是可变的。如果 access==1 它应该被命名为 Active ，否则它应该被命名为 Diactive.

这是我的代码和我得到的图片：

<?php
<table class="table table-condensed">
   <thead>
<tr>
  <th>ID</th>
  <th>User</th>
  <th>Operator</th>
  <th>View</th>
</tr>  
<?php
 while ($row = pg_fetch_array($result)) {
  ?>
<tr>
 <td>
  <?php
  $id = $row["id"];
  echo $id;
  ?>
 </td>
 <td>
  <?php
  $username = $row["username"];
  echo $username;
  ?>
 </td>
 <td>
  <form method="POST" action="oper.php">
<?php
include ("db.php");
 
$result2 = pg_query($db_connection, "SELECT * from users ORDER by id asc");
while ($row1 = pg_fetch_array($result2)) 
{
 $iddrain= $row1['id'];
 //echo $iddrain;
   
  
 //echo $iddrain;
 $q7= "Select access from nezeret where id_m=$iddrain";
 
 //var_dump($q7);
 $resultid= sqlsrv_query($link, $q7, $params, $options); 
 while($row7= sqlsrv_fetch_array($resultid))
 {
  //$rs7=$row7['ID_M'];
  $rs8=$row7['access'];
  //echo $rs8;
  //break;
 
 if($rs8==1)
  {
   echo "<p><input type=\"submit\" name=\"uid\" value=Operator-ON onchange=\"this.form.submit()\"></p>
    <p><input type=\"hidden\" name=\"uid\" value=$id onchange=\"this.form.submit()\"></p>"; 
  }
 
  else
  {
   echo "<p><input type=\"submit\" name=\"uid\" value=DIavtive onchange=\"this.form.submit()\"></p>
    <p><input type=\"hidden\" name=\"uid\" value=$id onchange=\"this.form.submit()\"></p>"; 
  }
 }
}
?>     
        </form>
 </td>
<?php
 }
?>



</tr>
</table>
?>

Answer 1

您犯了错别字，没有为附近的值属性提供报价：

像这样尝试：

 if($rs8==1)
    {
        echo '<p><input type="submit" name="uid" value="Operator-ON"  onchange="this.form.submit()"></p>
        <p><input type="hidden" name="uid" value="'.$id.'"  onchange=
    "this.form.submit()"></p>'; 
    }

   else
   {
        echo '<p><input type="submit" name="uid" value="DIavtive"  onchange="this.form.submit()"></p>
                 <p><input type="hidden" name="uid" value="'.$id.'  onchange="this.form.submit()"></p>'; 
        }
  }

Answer 2

我已经重新编辑了您的代码以删除导致它无法工作的一些错误；

<?php include ("db.php"); ?>    
        <table class="table table-condensed">
            <thead>
                <tr>
                    <th>ID</th>
                    <th>User</th>
                    <th>Operator</th>
                    <th>View</th>
                </tr>
            </thead>        
    <?php
 //while ($row = pg_fetch_array($result)) {

      <?php foreach( pg_fetch_array($result) as $row ) { ?>
     ?>
        <tbody>
            <tr>
                <td<?php echo $row['id'] ?></td>
                <td><?php echo $row['username'] ?></td>
                <td>
                    <form method="POST" action="oper.php">
    <?php 
        $result2 = pg_query($db_connection, 'SELECT * from users ORDER by id asc');
     while ($row1 = pg_fetch_array($result2)) {
         $iddrain = $row1['id'];
         $q7 = "Select access from nezeret where id_m=$iddrain";

        //var_dump($q7);
        $resultid = sqlsrv_query($link, $q7, $params, $options);
         while ($row7 = sqlsrv_fetch_array($resultid)) {
             //$rs7=$row7['ID_M'];
            $rs8 = $row7['access'];

            if ($rs8 == 1) {
                echo '<p><input type="submit" name="uid" value=Operator-ON onchange="this.form.submit()"></p>
                        <p><input type="hidden" name="uid" value=$id onchange="this.form.submit()"></p>';
            } else {
                echo '<p><input type="submit" name="uid" value=DIavtive onchange="this.form.submit()"></p>
                        <p><input type="hidden" name="uid" value=$id onchange="this.form.submit()"></p>';
            }
         }
     }  
     ?>     
                </form>
            </td>


        </tr>
        </tbody>
    </table>

Answer 3

我更改了代码，现在它可以正常工作了，问题是额外的无用提取。这是代码：

<table class="table table-condensed">
   <thead>
<tr>
  <th>ID</th>
  <th>User</th>
  <th>Operator</th>
  
</tr>  
<?php
 while ($row = pg_fetch_array($result)) {
  ?>
<tr>
 <td>
  <?php
  $id = $row["id"];
  echo $id;
  ?>
 </td>
 <td>
  <?php
  $username = $row["username"];
  echo $username;
  ?>
 </td>
 <td>
  <form method="POST" action="oper.php">
<?php
include ("db.php");
 
 $iddrain= $row['id'];

 $q7= "Select * from nezeret where id_m=$iddrain"; 
 //var_dump($q7);
 $resultid= sqlsrv_query($link, $q7, $params, $options); 
 while($row7= sqlsrv_fetch_array($resultid))
 {
  $rs8=$row7['access'];
  //echo $rs8;

 
  if($rs8==1)
  {
   echo "<p><input type=\"submit\" name=\"uid\" value=Operator onchange=\"this.form.submit()\"></p>
    <p><input type=\"hidden\" name=\"uid\" value=$id onchange=\"this.form.submit()\"></p>"; 
  }
 
  else
  {
   echo "<p><input type=\"submit\" name=\"uid\" value=Nazeret onchange=\"this.form.submit()\"></p>
    <p><input type=\"hidden\" name=\"uid\" value=$id onchange=\"this.form.submit()\"></p>"; 
  }
 }

?>     
        </form>
 </td>


</tr>
<?php
 }
?>

</tr>
</table>

为每一行打印提交按钮

Printing Submit button for each row

php

sql-server

input

submit

button