Question

我正在尝试使用summernote和airMode提交多个输入。当连接到＆＃34; textarea＆＃34;时，Airmode不起作用。这很烦人。如果可以，请告诉我。同时我试图破解所以我提交表单和我们PHP以向DB添加内容。

我可以提交一个＆＃34; p＆＃34;标签，但我需要传递多个字段。我需要＆＃34; id＆＃34;保留即$ _POST [p] [1]，$ _POST [p] [2]，$ _POST [p] [3]等。如果有更好的方法，请指出我正确的方向。

<?php echo print_r($_POST[p]); ?>
<form id="form" action="" method="POST" enctype="multipart/form-data" onsubmit="postForm()">
<p class="inline-editor" id="1">Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat.</p>
<p class="inline-editor" id="2">Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat.</p>
<p class="inline-editor" id="3">Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat.</p>

<input name="" type="submit">
</form>


<script type="text/javascript">
  var form = document.getElementById('form');
  form.onsubmit = function()
  {
    var p = this.getElementsByTagName('p')[0];
    if (!document.getElementById('pval'))
    {
      var pinput = document.createElement('input');
      pinput.setAttribute('type', 'hidden');
      pinput.setAttribute('id', 'pval');
      pinput.setAttribute('name', 'p');
      this.appendChild(pinput);
    }

    document.getElementById('pval').value = p.innerHTML;

    return true;
  }
</script>

Answer 1

向表单添加多个input - 元素，每个p - 元素一个元素，并将它们命名为p[1]，p[2]等：

<script type="text/javascript">
  var form = document.getElementById('form');
  form.onsubmit = function()
  {
    var pElems = this.getElementsByTagName('p');

    for(var i = 0; i < pElems.length; i++) {
        var p = pElems[i],
            id = p.id;

        if (!document.getElementById('pval_' + id))
        {
          var pinput = document.createElement('input');
          pinput.setAttribute('type', 'hidden');
          pinput.setAttribute('id', 'pval_' + id);
          pinput.setAttribute('name', 'p[' + id + ']');
          this.appendChild(pinput);
        }

        document.getElementById('pval_' + id).value = p.innerHTML;
    }

    return true;
  }
</script>

Answer 2

if summernote - bootstrap - jQuery jsfiddle

$query  = "SELECT external_source_id FROM smf_tds_unique_statistic WHERE ".$dateRange." GROUP BY external_source_id ORDER BY external_source_id ASC;";
$query .= "SELECT sources_id FROM smf_tds_unique_statistic WHERE ".$dateRange." GROUP BY sources_id ORDER BY sources_id ASC;";
$query .= "SELECT ip_country FROM smf_tds_unique_statistic WHERE ".$dateRange." GROUP BY ip_country ORDER BY ip_country ASC;";
$query .= "SELECT cidr FROM smf_tds_unique_statistic WHERE ".$dateRange." GROUP BY cidr ORDER BY cidr ASC;";

if (mysqli_multi_query($db, $query)) {
do {
    if ($result = mysqli_store_result($db)) {

        while ($row = mysqli_fetch_row($result)) {
            $val = $row[0];
            if ($val != '') {
                $options .= '<li class="select_input_option_item" value="'.$val.'">'.$val.'</li>';
            }
        }
        array_push($optionsArr, $options);
        $options = '';

        mysqli_free_result($result);
    }

    if (!mysqli_more_results($db)) break;
} while (mysqli_next_result($db));} mysqli_close($db);

使用summernotes提交<p>标签

2 个答案: