passing checkbox ID to another php file

问题

let's say i'm generating some sort of html table (from mysql queried data) with a checkbox at each row with something like

$display_string = "<form action=\"delete.php\" method=\"post\">";
$display_string .= "<div><input type=\"button\" VALUE=\"Button1\"></div>";
$display_string .= "<div><input type=\"button\" VALUE=\"Button2\"></div>";
while($row = mysql_fetch_array($qry_result)){
    $display_string .= "<tr onmouseover=\"this.className = 'hlt';\" onmouseout=\"this.className = '';\">";
    $display_string .= "<td class='blank'><input type=\"checkbox\" /></td>";
    $display_string .= "<td class='data'>" . $row['first_name'] . "</td>";
    $display_string .= "<td class='data'>" . $row['last_name'] . "</td>";
    $display_string .= "<td class='data'><a href='" . $row['email'] . "'>" . $row['email'] . "</a></td>";
etc...
etc...
$display_string .= "</form>";

what i'd like to happen now, after various checkboxes are selected are two things: 1) for example, call delete.php when Button1 is clicked to delete the selected rows. 2) some other php file called when Button2 is clicked

I have access to $row['ID'] which I can use to name each checkbox, i'm just not sure how to incorporate it since i'm new to php

---

UPDATE The following seems to work for my purposes

I've got the following html-

<form name='myForm' method=\"post\" >
<input type=\"submit\" onClick=\"deleterow(document.myForm)\" VALUE=\"Delete ROWs\">

while($row = mysql_fetch_array($qry_result)){
    <input type=\"checkbox\" name='rows' value=" .$row['ID']. "/>

Javascript is as follows-

<script language=\"javascript\" type=\"text/javascript\">
function deleterow(form){

    if (!confirm(\"Are you sure you want to delete?\")) return false;

    var queryString = \"?ID=\";

    for (var i = 0; i < document.myForm.rows.length; i++) {
        if (document.myForm.rows[i].checked) {
            ID = document.myForm.rows[i].value;
            ID = ID.slice(0, -1);
            queryString += ID;
            queryString += \"-\";
        }
    }
    queryString = queryString.slice(0, -1);

    try{
        // Opera 8.0+, Firefox, Safari
        ajaxRequest = new XMLHttpRequest();
    } catch (e){
        // Internet Explorer Browsers
        try{
            ajaxRequest = new ActiveXObject(\"Msxml2.XMLHTTP\");
        } catch (e) {
            try{
                ajaxRequest = new ActiveXObject(\"Microsoft.XMLHTTP\");
            } catch (e){
                // Something went wrong
                alert(\"Your browser broke!\");
                return false;
            }
        }
    }

    var ajaxRequest;  // The variable that makes Ajax possible!
    // Create a function that will receive data sent from the server
    ajaxRequest.onreadystatechange = function(){
        if(ajaxRequest.readyState == 4){
            var ajaxDisplay = document.getElementById('ajaxDiv');
            ajaxDisplay.innerHTML = ajaxRequest.responseText;
        }
    }

    ajaxRequest.open(\"GET\", \"delete_row.php\" + queryString, true);
    ajaxRequest.send(null); 
    confirm('Delete successful!');
}

then with delete_row.php you can do all you need to with a mysql query, and new data can be sent back and displayed with

<div id='ajaxDiv'></div>

回答1:

To get a clear code I suggest you to avoid two buttons firing two several actions. Instead use a simple javascript method to append a parameter to the submitted form to get rid of what button has been pushed.

<script type="text/javascript">
    function submitForm( act ){
         $('#act').val(act);
         $('#myForm').submit();
}

</script>

<form id="myForm" ... >
<input type="hidden" id="act" value="" />

<button onClick="submitForm(1)">1</button>
<button onClick="submitForm(2)">2</button>
</form>

this is just an example (using jquery).

About your second answer (checkboxes):

<input type="checkbox" name="selectedboxes[]" value="someIdHere" />

when the form is submitted you'll receive an array "selectedboxes" as a param in which you get checked checkboxes.

回答2:

You can do it changing the Form action, for example if you click BUTTON 1 it will change the action to DELETE.PHP, take a look to this post:

Changing the action of a form with javascript/jquery

They use JQuery and they have multiples answer how to do it

来源：https://stackoverflow.com/questions/11119719/passing-checkbox-id-to-another-php-file