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我已经阅读了所有具有相同问题但没有成功的问题。
我试过trigger(), unbind('click'), on('click', function()), live('click'), 等等

我的问题是我必须在保存按钮上单击两次以验证并将数据提交到数据库并返回消息'success'或'fail'以显示错误消息。

查询

$('.save').click(function (event) {
    // GET CATEGORIA
    var category_change = 0;
    if (!$('#new_category').val()) {
        category = $('#categoria_id').val();
    } else {
        category = $('#new_category').val();
        category_change = 1;
    }

    // GET TITLE
    var title = $('input#title').val();

    // GET IMAGE
    var target = $('#parent');
    html2canvas(target, {
        onrendered: function (canvas) {
            image = canvas.toDataURL();
        }
    });

    // GET FORMATED CODE
    var array = ["n", "e", "s", "w"];
    var format = $('#parent').html();
    var strrep = '<div class="close">X</div>';
    var strrep2 = '<div class="ui-resizable-handle ui-resizable-se ui-icon ui-icon-gripsmall-diagonal-se" style="z-index: 1000;"></div>';

    var aux2 = 0;
    while (aux2 < 5) {
        $.each(array, function () {
            format = format.replace('<div class="ui-resizable-handle ui-resizable-' + this + '" style="z-index: 1000;"></div>', '');
        });
        format = format.replace(strrep, '');
        format = format.replace(strrep2, '');
        aux2++;
    }
    var code = '<div id="parent">' + format + '</div>';


    if (title != '' && $('.child').length != 0) {
        $('.db_result').html('<img src="images/loader.gif" />');
        $.post("db_prox.php", {
            category: category,
            title: title,
            image: image,
            code: code,
            category_change: category_change
        }, function (post_result) {
            alert('2');
            if (post_result == 'success') {
                $('.db_result').text('O template foi guardado com sucesso!');
                $('.continue').prop("disabled", false);
            } else {
                $('.db_result').text('Ocorreu um erro na gravação do template!');
            }
        });
    }

    // INSERIR NA BASE DE DADOS CLOSE
});

PHP db_prox.php

<?php
require_once 'includes/functions.php';

$categoria = $_POST['category'];
$title = $_POST['title'];
$image = $_POST['image'];
$code = $_POST['code'];
$nova_categoria = $_POST['category_change'];

if($nova_categoria == 1){

    $result = get_cat_last_created();
    $cat_id = mysqli_fetch_assoc($result);
    $aux_id = $cat_id['categoria_id'] + 1;
    $query = "INSERT INTO categorias (categoria_id, name) VALUES ('$aux_id', '$categoria')";
    if(mysqli_query($connection, $query)) {
        $query2 = "INSERT INTO templates (categoria_id, title, image, code) VALUES ('$aux_id', '$title', '$image', '$code')";

        if(mysqli_query($connection, $query2)) {
            echo "success";
        }
        else {
            echo "fail";
        }
    }
}else{
    $query = "INSERT INTO templates (categoria_id, title, image, code) VALUES ('$categoria', '$title', '$image', '$code')";
    if(mysqli_query($connection, $query)) {
        echo "success";
    }
    else {
        echo "fail";
    }
}
mysqli_close($connection);
?>
4

1 回答 1

1

该onrendered函数是异步的

 html2canvas(target, {
    onrendered: function (canvas) {
        image = canvas.toDataURL();
    }
});

这意味着在此之后同步运行的任何代码都没有image分配变量。

在函数内移动你的 ajax 帖子onrendered将使它在创建图像后发送请求，即：

 html2canvas(target, {
    onrendered: function (canvas) {
        image = canvas.toDataURL();
        $.post("db_prox.php", {
          category: category,
          title: title,
          image: image,
          code: code,
          category_change: category_change
        }, function (post_result) {
          alert('2');
          if (post_result == 'success') {
            $('.db_result').text('O template foi guardado com sucesso!');
            $('.continue').prop("disabled", false);
          } else {
            $('.db_result').text('Ocorreu um erro na gravação do template!');
          }
        });
    }
});
于 2013-08-13T17:09:15.977 回答