小编Zen*_*eng的帖子

我试图将以下图像绘制到画布上,但尽管定义了画布的大小,但它看起来很模糊.正如您在下面看到的,图像清晰而清晰,而在画布上,图像模糊且像素化.

这是它的外观(左边是原始的,右边是绘制的画布,模糊不清.)

我究竟做错了什么？

console.log('Hello world')

var c = document.getElementById('canvas')
var ctx = c.getContext('2d')
var playerImg = new Image()

// http://i.imgur.com/ruZv0dl.png sees a CLEAR, CRISP image
playerImg.src = 'http://i.imgur.com/ruZv0dl.png'
playerImg.width = 32
playerImg.height = 32

playerImg.onload = function() {
  ctx.drawImage(playerImg, 0, 0, 32, 32);
};

#canvas {
  background: #ABABAB;
  position: relative;
  height: 352px;
  width: 512px;
  z-index: 1;
}

<canvas id="canvas" height="352" width="521"></canvas>

无论出于何种原因,我browserify并gulp停止工作.例如,这是我的gulp js脚本捆绑我的JavaScript.

gulp.task('js', function() {
    gulp.src('src/js/*.js')
        .pipe(plumber())
        .pipe(gulp.dest('js'));
    return browserify('./src/js/main', { debug: true })
    .bundle()
    .pipe(source('bundle.js'))
/*    .pipe(streamify(uglify()))*/
    .pipe(gulp.dest('.'))
    .pipe(notify({ message: 'JavaScript has been bundled with Browserify!'}));
    // .pipe(livereload());
});

这是main.js:

var ajaxChng = require('./ajax-changelog');
ajaxChng();

里面src/js/ajax-changelog.js是这样的:

module.exports = {
    console.log('Hello World');
};

但当我这样做时gulp js,我得到:

? gulp js
[19:11:50] Using gulpfile c:\wamp\www\osrsmap\gulpfile.js
[19:11:50] Starting 'js'...

events.js:85
      throw er; // Unhandled 'error' event
            ^
SyntaxError: Unexpected token

... 我究竟做错了什么？

我一直在讨论这个MySQL查询.

假设我有这个:

    $add = "INSERT INTO books (title) VALUES(?)";
    if ($stmt = $mysqli->prepare($add)) {

        $arr = array($title);

        foreach ($arr as $value) {
            echo var_dump($value);
        }

        $stmt->bind_param("s", $title);

有了foreach- > var_dump:

string 'Medieval Times (History)' (length=24)
int 1422843281
int 1420844341
string '127.0.0.1' (length=9)
string 'MY_EMAIL@gmail.com' (length=22)
string '' (length=0)
int 1420844805
int 6
int 3
int 1
int 0
int 0
int 1
int 1
int 1
int 1

好吧,当它到达这一行时它会停止并且我收到此错误:

Fatal error: Call to a member function bind_param() …

所以,我有这个函数可以获得一个JSON对象,但我想让它更简单,所以我创建了一个函数来获取JSON对象的值.为什么不起作用？

var itemData = {
    weapon: function () {
        return {
            1: {
                'name': 'Dagger',
                    'extra_skill': 'none',
                    'cost': 500,
                    'attack': 5
            },
            2: {
                'name': 'Pickaxe',
                    'extra_skill': 'mining',
                    'cost': 25,
                    'attack': 5
            }
        }
    },
    getWeapon: function (value, x) {
        var obj = JSON.parse(value);
        return itemData.weapon()[x].obj
    }
}

// outputs: Dagger
console.log(itemData.weapon()[1].name)

// Get the name of weapon 1
// however, it outputs: Uncaught SyntaxError: Unexpected token a
console.log('Getting weapon... ' + itemData.getWeapon('name', 1))

我究竟做错了什么？