如果复选框被选中，则显示当前行的成功图像

Question

我有这样的代码:

<script>
$(document).ready(function() {
    $("#div_trattamenti").on('change', '.inpt_chkbox_trattamenti', function (e) {
        e.preventDefault();
        if($(this).is(":checked")){
        var id =  $(this).attr('rel');
        var value = $(this).val();
        alert(value);                                                                                                                           // prezzo
        }if($(this).is(":not(:checked)")){
            var id =  $(this).attr('rel');
            var value = 0;
            alert(value);                                                                                                                           // prezzo
        }
        data_chk = 'id=' + id + '&value=' + value;
        $.ajax({
            type: "POST",
            data: data_chk,
            cache: false,
            url: "php/aggiorna_chiuso.php",
            success: function(html){
                               var obj = $.parseJSON(html);
                                $.each(obj,function(index, element){
                                    var value2 = element.chiuso;                                // value chiuso
                                    }); // fine .each
                                if(value == 1){
                                    $('.chkmk_img_trattamenti', this).show();
                                }
                                if(value == 0){
                                    $('.chkmk_img_trattamenti', this).hide();
                                }
            },                                                                                  // fine success
            error: function(xhr, status, error) {
                console.log(status);
                console.log(error);
                console.dir(xhr);
                }                                                                       // fine error
            });                                                                         // fine ajax
    });
});
</script>

它的工作，但我不能显示图像仅为选定的行(当用户单击复选框)。

我试过这个，但它不工作。:

$('。chkmk_img_trattamenti’,这个),告诉();

html代码:

    $html = '';                         
    $html .= "<table class='class_table'>";
while( $row = mysql_fetch_array($query_1)){
    $html .= "<tr class='row'>";
    $html .= "<td class='col_1'>";
    $html .= "$row[prestazione]";
    $html .= "</td>";
    $html .= "<td class='col_2'>";
    $html .= "$row[denti]";
    $html .= "</td>";
    $prezzo_2 = number_format($row[prezzo], 0, ',', '.');
    $html .= "<td class='col_1'>";
    $html .= "$prezzo_2";
    $html .= "</td>";       
    $html .= "<td class='col_3'>";
    $html .= "<a href='' class='a_mod_t' rel='$row[id]'><img src='icons/mod.png' border='0' title='modifica trattamento'/></a>";
    $html .= "</td>";
    $html .= "<td class='col_3'>";
    $html .= "<a href='' class='a_mod_prezzo' rel='$row[id]'><img src='icons/euro.png' border='0' title='modifica prezzo'/></a>";
    $html .= "</td>";           
    $html .= "<td class='col_3'>";
    $html .= "<a href='' class='a_delete_t' rel='$row[id]'><img src='icons/delete.png' border='0' title='cancella trattamento'/></a>";
    $html .= "</td>";
    $html .= "<td class='col_4'>";
    $html .= "<div class='chkbox_trattamenti'>";
    $html .= "<form class='form_chkbox_trattamenti'>";
    $html .= "<input type='checkbox' class='inpt_chkbox_trattamenti' value='1' rel='$row[id]'>";
    $html .= "<input class='hidden_id_dottore_chk' type='hidden' value='<?php echo $id; ?>''>";
    $html .= "<input class='hidden_id_paz_chk' type='hidden' value='<?php echo $paziente_id; ?>'>";
    $html .= "<input class='hidden_id_chk' type='hidden' value='<?php echo $trattamento_id; ?>'>";
    $html .= "</form>";
    $html .= "<div class='chkmk_img_trattamenti'>";
    $html .= "<img src='icons/checkmark.png' border='0' title='trattamento chiuso' />";
    $html .= "</div>";
    $html .= "</div>";
    $html .= "</td>";

Answer 1

您需要获得currentdiv的对象。

创建一个名为successMsg的变量，并将当前对象赋值给它，如下所示

var successMsg = $(this);

并在成功方法中调用它，像这样$(successMsg).parent().parent().find('.chkmk_img_trattamenti').show();

这是更新后的JS

$(document).ready(function() {
    $("#div_trattamenti").on('change', '.inpt_chkbox_trattamenti', function(e) {
        e.preventDefault();
        if ($(this).is(":checked")) {
            var id = $(this).attr('rel');
            var value = $(this).val();
            alert(value); // prezzo
        }
        if ($(this).is(":not(:checked)")) {
            var id = $(this).attr('rel');
            var value = 0;
            alert(value); // prezzo
        }
        var successMsg = $(this);
        data_chk = 'id=' + id + '&value=' + value;
        $.ajax({
            type: "POST",
            data: data_chk,
            cache: false,
            url: "php/aggiorna_chiuso.php",
            success: function(html) {
                var obj = $.parseJSON(html);
                $.each(obj, function(index, element) {
                    var value2 = element.chiuso; // value chiuso
                }); // fine .each
                if (value == 1) {
                    $(successMsg).parent().parent().find('.chkmk_img_trattamenti').show();
                    //$('.chkmk_img_trattamenti', this).show();
                }
                if (value == 0) {
                    //$('.chkmk_img_trattamenti', this).hide();
                    $(successMsg).parent().parent().find('.chkmk_img_trattamenti').hide();
                }
            }, // fine success
            error: function(xhr, status, error) {
                    console.log(status);
                    console.log(error);
                    console.dir(xhr);
                } // fine error
        }); // fine ajax
    });
});

Answer 2

    $html = '';                         
    $html .= "<table class='class_table'>";
while( $row = mysql_fetch_array($query_1)){
    $html .= "<tr class='row'>";
    $html .= "<td class='col_1'>";
    $html .= "$row[prestazione]";
    $html .= "</td>";
    $html .= "<td class='col_2'>";
    $html .= "$row[denti]";
    $html .= "</td>";
    $prezzo_2 = number_format($row[prezzo], 0, ',', '.');
    $html .= "<td class='col_1'>";
    $html .= "$prezzo_2";
    $html .= "</td>";       
    $html .= "<td class='col_3'>";
    $html .= "<a href='' class='a_mod_t' rel='$row[id]'><img src='icons/mod.png' border='0' title='modifica trattamento'/></a>";
    $html .= "</td>";
    $html .= "<td class='col_3'>";
    $html .= "<a href='' class='a_mod_prezzo' rel='$row[id]'><img src='icons/euro.png' border='0' title='modifica prezzo'/></a>";
    $html .= "</td>";           
    $html .= "<td class='col_3'>";
    $html .= "<a href='' class='a_delete_t' rel='$row[id]'><img src='icons/delete.png' border='0' title='cancella trattamento'/></a>";
    $html .= "</td>";
    $html .= "<td class='col_4'>";
    $html .= "<div class='chkbox_trattamenti'>";
    $html .= "<form class='form_chkbox_trattamenti'>";
    $html .= "<input type='checkbox' class='inpt_chkbox_trattamenti' value='1' rel='$row[id]'>";
    $html .= "<input class='hidden_id_dottore_chk' type='hidden' value='<?php echo $id; ?>''>";
    $html .= "<input class='hidden_id_paz_chk' type='hidden' value='<?php echo $paziente_id; ?>'>";
    $html .= "<input class='hidden_id_chk' type='hidden' value='<?php echo $trattamento_id; ?>'>";
    $html .= "</form>";
    $html .= "<div class='chkmk_img_trattamenti'>";
    $html .= "<img src='icons/checkmark.png' border='0' title='trattamento chiuso' />";
    $html .= "</div>";
    $html .= "</div>";
    $html .= "</td>";

Answer 3

隐藏/显示元素:

$("#div_trattamenti").on('change', '.inpt_chkbox_trattamenti', function (e) {
       .
       .
       .
      $("#" + this.id).find(" .chkmk_img_trattamenti").show();
       .
       .
       .
});