html javascript의 버튼이 작동하지 않습니다.

Question

PHP 스크립트를 실행하기 위해 양식을 제출하려고합니다. 하지만 제출 버튼이 작동하지 않습니다. 따옴표 나 표현이 빠진 것 같지 않습니다. 내가 뭘 잘못하고 있니?

<script type="text/javascript"> 

function doSubmit() 
{ 

<?php 
echo "Hello"; 
?> 
var actions = document.frm.actions.value; 
var tables = document.frm.tables.value; 
var text0 = document.frm.text0.value; 
var text1 = document.frm.text1.value; 
var text2 = document.frm.text2.value; 

/*var txtBox = document.getElementById("text3"); 

if (txtBox!=null){ 
var text3 = document.frm.text3.value; 
var strURL = "devicesql.php?actions=" +actions+ "&tables=" +tables+ "&text0=" +text0+  "&text1=" +text1+ "&text2=" +text2+ "&text3=" +text3; 
} 

else 
{*/ 
var strURL = "devicesql.php?actions=" +actions+ "&tables=" +tables+ "&text0=" +text0+ "&text1=" +text1+ "&text2=" +text2; 
//} 

var xmlhttp; 

if (str=="") 
    { 
    document.getElementById("result").innerHTML=""; 
    return; 
    } 
if (window.XMLHttpRequest) 
    { 
    xmlhttp=new XMLHttpRequest(); 
    } 
else 
    { 
    xmlhttp=new ActiveXObject("Microsoft.XMLHTTP"); 
    } 
xmlhttp.onreadystatechange=function() 
    { 
    if (xmlhttp.readyState==4 && xmlhttp.status==200) 
{ 
document.getElementById("result").innerHTML=xmlhttp.responseText; 
} 
    } 

xmlhttp.open("GET",strURL,true); 
xmlhttp.send(null); 

} 

</script> 

<form name="frm" method="get" action="doSubmit();"> 


<?php 

// Connect to server and select database. 
mysql_connect("127.0.0.1:3306", "root", "")or die("cannot connect"); 
mysql_select_db("PushApplication")or die("cannot select DB"); 

$sql="SELECT * FROM Device"; 
$result=mysql_query($sql); 

$count=mysql_num_rows($result); 

?> 
<table width="400" border="0" cellspacing="1" cellpadding="0"> 
<tr> 
<td><form name="form1" method="post" action=""> 
<table width="400" border="0" cellspacing="1" bgcolor="#CCCCCC"> 

<tr> 
<td align="center" bgcolor="#FFFFFF">#</td> 
<td align="center" bgcolor="#FFFFFF"><strong>ID</strong></td> 
<td align="center" bgcolor="#FFFFFF"><strong>DeviceID</strong></td> 
<td align="center" bgcolor="#FFFFFF"><strong>DeviceType</strong></td> 
<td align="center" bgcolor="#FFFFFF"><strong>Description</strong></td> 
<td align="center" bgcolor="#FFFFFF"><strong>OS_Version</strong></td> 
<td align="center" bgcolor="#FFFFFF"><strong>Status</strong></td> 

</tr> 
<?php 
while($rows=mysql_fetch_array($result)){ 
?> 
<tr> 
<td align="center" bgcolor="#FFFFFF"><input name="checkbox[]" type="checkbox" id="checkbox[]" value="<? echo $rows['ID']; ?>"></td> 
<td bgcolor="#FFFFFF"><? echo $rows['ID']; ?></td> 
<td bgcolor="#FFFFFF"><? echo $rows['DeviceID']; ?></td> 
<td bgcolor="#FFFFFF"><? echo $rows['DeviceType']; ?></td> 
<td bgcolor="#FFFFFF"><? echo $rows['Description']; ?></td> 
<td bgcolor="#FFFFFF"><? echo $rows['OS_Version']; ?></td> 
<td bgcolor="#FFFFFF"><? echo $rows['Status']; ?></td> 
</tr> 
<?php 
} 
?> 
<tr> 
<td colspan="7" align="center" bgcolor="#FFFFFF"><input name="delete" type="submit" id="delete" value="Delete"></td> 
</tr> 
<?php 
// Check if delete button active, start this 
if($delete){ 
for($i=0;$i<$count;$i++){ 
$del_id = $checkbox[$i]; 
$sql = "DELETE FROM Device WHERE ID='$del_id'"; 
$result = mysql_query($sql); 
} 

// if successful redirect to delete_multiple.php 
if($result){ 
echo "<meta http-equiv=\"refresh\" content=\"0;URL=delete_multiple.php\">"; 
} 
} 
mysql_close(); 
?> 
</table> 
</form> 
</td> 
</tr> 
</table> 

<br /> 
<br /> 


Action: <id="actions"> 
<select name="actions"> 
<option value="insert into ">Insert</option> 
<option value="update">Update</option> 
<option value="select">Display</option> 
</select> 

</br> 
</br> 

Database Table Name: <id="tables"> 
<select name="tables" onchange="addRows(this.value)"> 
<option>Select Table</option> 
<option value="Applications">Application</option> 
<option value="Device">Device</option> 
</select> 
</br> 

<div id="addedRows"></div> 

</br> 

<input type="submit" value="Submit" onclick="doSubmit()"> 

</br> 
</br> 

Result: <div id="result"></div> 

</form> 

Answer 1

에 시도를 만드는 데이터베이스 정보를 표시하는 테이블의 다른 양식을 작성하십시오. 양식이 2 개의 버튼과 혼동 될 수 있습니다.

Answer 2

당신은 양식에서 동작 특성을 제거해야하고, 양식을 제출 입력이 함수를 발생하고 이벤트를 제출하지 않은 온 클릭 속성이 있어야합니다.

<form id="theform"> 
    <!-- some input texts.... or something --> 
    <button onclick="processForm()">Process form</button> 
</form> 

<script type="text/javascript"> 

    function processForm() { 
     var form = document.getElementById('theform'); 

     // your stuff 
    } 

</script> 

PS : 당신은

Answer 3

사용이 양식을 제출하는 대신 버튼을 제출하는 코드를 다음에 "Hello"에코 첫 PHP 태그와 sintax 오류,

<a href="javascript: doSubmit()">Submit</a>