<p align="left">
<b>Click this button to create div element dynamically:</b>
<?php $qry = "SELECT COUNT(*) AS count From Contact Where CustomerID=$pid";
$result = mysql_query($qry);
$row = mysql_fetch_array($result);
$count = $row['count'];
echo $count;
if($count >= 6)
{
?>
<input id="btn1" type="button" value="create div" onClick="popup();" />
<?php
}
else
{
$num = 6 - $count ;
echo $num;
?>
<input id="btn1" type="button" value="create div" onClick="createDiv(<?php echo $num ?>);" />
<?php
}
?>
我想以「$ NUM」傳遞給我創造DIV功能如何將我的PHP中的值傳遞給javascript函數?
如果$ num是6也可以創建6格,如果$ num是4只生成多達4個格只
這個我createDiv功能
var i=0;
function createDiv(num)
{
if(i < num) {
var divTag = document.createElement("div");
divTag.id = "div1"+i;
divTag.setAttribute("align","left");
divTag.style.margin = "0px auto";
divTag.className ="ex";
divTag.innerHTML = "<img class='myimage' onclick='changeimage(this)' border='0' src='images/white_contact.png' width='60'/><table border='0'><tr><td>Name:</td><td><input type='text'></d></tr><tr><td>Title:</td><td><input type='text'></td></tr><tr><td>Contact:</td><td><input type='text'></td></tr></table>";
document.getElementById("newdiv").appendChild(divTag)
}
i++;
$(".ex").draggable({containment:'parent',cursor:'pointer',opacity:0.6, });
$(".ex").droppable({ hoverClass:'border' });
}
但現在我只能創建一個DIV,爲什麼會這樣?
幾乎相同的代碼。 – Asenar