我正在尝试将信息从动态表单传递到我的数据库。我正在使用jquery为表单中的数量和成分添加其他字段。然后,我希望能够将这些详细信息提交到我的数据库。 jQuery工作正常,但我不确定如何使用PHP来做到这一点。我正在使用以下代码:
HTML
<form id="htmlForm" method="post" action="includes/add-recipe.php" class="form-inline">
<input type="text" name="recipeName" class="input-large" placeholder="Recipe Title"><br /><br />
<div id="input1" style="margin-bottom:4px;" class="clonedInput">
Ingredient:<br />
<select type="text" name="entry[][$ing]"/>
<option value="Eggs">Eggs</option>
<option value="Flour">Flour</option>
<option value="Wheat">Wheat</option>
</select><input type="text" name="entry[][$qty]" class="input-small" placeholder="Quantity">
</div><br />
<div>
<input type="button" id="btnAdd" class="input-small" value="add another ingredient" />
<input type="button" id="btnDel" class="input-small" value="remove name" />
</div><br />
<button type="submit" class="btn">Add Recipe</button>
</form>
的Javascript
<script type="text/javascript">
$(document).ready(function() {
$('#btnAdd').click(function() {
var num = $('.clonedInput').length; // how many "duplicatable" input fields we currently have
var newNum = new Number(num + 1); // the numeric ID of the new input field being added
var newElem = $('#input' + num).clone().attr('id', 'input' + newNum);
newElem.children(':first').attr('id', 'entry' + newNum).attr('entry', 'entry' + newNum);
$('#input' + num).after(newElem);
$('#btnDel').attr('disabled','');
if (newNum == 5)
$('#btnAdd').attr('disabled','disabled');
});
$('#btnDel').click(function() {
var num = $('.clonedInput').length; // how many "duplicatable" input fields we currently have
$('#input' + num).remove(); // remove the last element
$('#btnAdd').attr('disabled','');
if (num-1 == 1)
$('#btnDel').attr('disabled','disabled');
});
$('#btnDel').attr('disabled','disabled');
});
</script>
PHP
<?php
foreach ($_POST["entry"] AS $aQtyIng)
{
list($ing, $qty) = $aQtyIng;
echo $aQtyIng;
}
?>