将多个值（文件）从javascript发送到php

Question

I have this code

var action = $("#action").val();
var z_product_id = $("#z_product_id").val(); 
var product_id = $("#product_id").val(); 
$.post('ajaxGetOperations.php',
    {action: action, z_product_id: z_product_id, product_id: product_id},
    function(data)
    {
        $('#respons').html(data);
        $('#hand_product_id').val(product_id);
    }) ;                                         

}

I can send the "action" and "Z_product" and ... in this line to php page

{action: action, z_product_id: z_product_id, product_id: product_id}

but I don't know what do I do for send file value to php page.
this is may html code

<table class="table table-striped table-hover">  
    <tr>
        <td><input type='text' size="4" value='$z_product[id]' id='z_product_id' /></td>
        <td>$z_product[name]</td>                          
        <td><input type='text' size="4" value='$product[id]' id='product_id' /></td>
        <td>$product[name]</td>                                                              
    </tr> 
    <tr>      
        <td colspan='4'>
            <div class="col-lg-6 col-sm-6 col-md-6 -col-xs-12">
                <label class="" for="filebutton">ارسال عکس</label>
                <input id="image" name="image" class="input-file" type="file">
            </div>
        </td>                                                               
    </tr>                                                               
    </tr>         
    <tr>      
        <td></td>
        <td></td>     
        <td></td>     
        <td>
            <button onclick='setDBInfo(this.value)' value='1' id='action'>ارسال </button>
            <button onclick='setDBInfo(this.value)' value='0' id='action'>رد </button>
        </td>                                                               
    </tr>
</table> 

actually I couldn't access the file value in php.
I want to access multiple file value in php whit javascript .post?
how can I do this?

Answer 1

The easiest way is to wrap your table in a form and send it to the php page. But if you don't want to do it that way, then you'll need to write js code to fetch the file value from the HTML element.

Answer 2

Try This

<form name="my_form">
<table class="table table-striped table-hover">  
    <tr>
        <td><input type='text' size="4" name="z_product_id" value='$z_product[id]' id='z_product_id' /></td>
        <td>$z_product[name]</td>                          
        <td><input type='text' size="4" name="product_id" value='$product[id]' id='product_id' /></td>
        <td>$product[name]</td>                                                              
    </tr> 
    <tr>      
        <td colspan='4'>
            <div class="col-lg-6 col-sm-6 col-md-6 -col-xs-12">
                <label class="" for="filebutton">ارسال عکس</label>
                <input id="image" name="image" class="input-file" type="file">
            </div>
        </td>                                                               
    </tr>                                                               
    </tr>         
    <tr>      
        <td></td>
        <td></td>     
        <td></td>     
        <td>
            <button type="submit" value='1' id='action'>ارسال </button>
            <button onclick='setDBInfo(this.value)' value='0' id='action'>رد </button>
        </td>                                                               
    </tr>
</table> 
</form>

and replace your ajax request with this

$("form[name='my_form']").submit(function(e) {
    var formData = new FormData($(this)[0]); /* this is your data */

        $.ajax({
            url: "ajaxGetOperations.php",
            type: "POST",
            data: formData,
            async: false,
            success: function (data) {
                $('#respons').html(data);
                $('#hand_product_id').val(product_id);
            },
            cache: false,
            contentType: false,
            processData: false
        });

           e.preventDefault();
        });

in your php retrieve the values as usual

$p_id = $_POST['z_product_id'];
$p_id = $_FILES['image'];

In short using the formData object would really help you...