Question

我想在控制器内回显一个表单。在那我写了echo form_open，但它不起作用。这是我在控制器函数中的代码

 echo '<button class="btn btn-primary" data-toggle="modal" style="background-color:#428bca;"  data-target="#myModal" ></button>';

                            echo '<div class="modal fade bs-example-modal-sm" tabindex="-1"  id="myModal"  role="dialog" aria-labelledby="mySmallModalLabel" aria-hidden="true">';
                              echo '<div class="modal-dialog modal-sm">';
                                echo '<div class="modal-content">';
                                   echo ' <div class="modal-header">';
                                        echo '<button aria-hidden="true" data-dismiss="modal"  class="close" type="button">&times;</button>';
                                        echo '<h4> Do you </h4>';?>

                                        <?php echo  form_open('amt_controller/add_amount/'.$schoolID."/".$stID);

                                          echo '<input type="name" class="form-control"  name="Amount" id="Amount" placeholder="Enter the amount">';

                                        echo '</div>';
                                      echo '<div class="modal-body">';
                                        echo '<button class="btn btn-primary" style="margin-left:65px;background-color:#428bca;">Submit</button>';?>
                                             <?php echo form_close();
                                       echo '<button type="button" class="btn btn-default" data-dismiss="modal">No</button>';
                                echo '</div>';
                                echo '</div>';
                              echo '</div>';
                            echo '</div>';

Answer 1

删除表单上的php标签打开和关闭，并将它们放置为

echo <<<_END
<html><head><title>PHP from Upload</title></head><body>
<form method = 'POST' action='upload.php' enctype='multipart/form.data'>
Select a JPG, GIF, PNG or TIF File: <input type='file' name='filename' size='10'>
<input type='submit' value='Upload'>
</form>
_END;

if ($_FILES){
    $name = $_FILES['filename']['name'];

    switch($_FILES['filename']['name']){
        case 'image/jpeg': $ext = 'jpg'; break;
        case 'image/gif': $ext = 'gif'; break;
        case 'image/png': $ext = 'png'; break;
        case 'image/tiff': $ext = 'tif'; break;
        default: $ext = ''; break;
    }
    if ($ext){
        $n = "image.$ext";
        move_uploaded_file($_FILES['filename']['tmp_name'], $n);
        echo "<img src='$n'>";
    }
    else echo "'$name' is not an accepted image file";
}
else echo "No image has been uploaded";
echo "</body></html>";

form_open无法在codeigniter控制器内部工作

1 个答案: