我正在使用instafeed js从Instagram调用照片。有没有办法将每4个图像包装在div中?这有可能吗?这是我的代码:
jQuery(window).load(function(){
var userFeed = new Instafeed({
get: 'user',
userId: 1509441625,
accessToken: '1509441625.5b9e1e6.c20b3eb91b15404ab30084283ab3a9c9',
limit : 4,
resolution:'standard_resolution',
template: '<a target="_Blank" href="{{link}}"><img src="{{image}}" /></a> ',
});
userFeed.run();
});
我联系了instafeed的开发人员,他给了我一个未经测试的代码,我试图调试:
var count = 0;
var userFeed = new Instafeed({
get: 'user',
userId: 1509441625,
accessToken: '1509441625.5b9e1e6.c20b3eb91b15404ab30084283ab3a9c9',
limit : 4,
resolution:'standard_resolution',
filter: function(image) {
count += 1;
if (count % 4 === 0) {
image.customTagOpen = '<div>';
image.customTagClose = '</div>';
} else {
image.customTagOpen = '';
image.customTagClose = '';
}
return true;
},
template: '{{model.customTagOpen}}<a target="_Blank" href="{{link}}"><img src="{{image}} /></a>{{model.customTagClose}}';
});
userFeed.run();
});
但是我收到错误:在列出房产后遗漏了“}”。有什么建议吗?
答案 0 :(得分:3)
template
选项行的末尾有一个错误放置的分号。
尝试将分号;
更改为逗号,
template: '{{model.customTagOpen}}<a target="_Blank" href="{{link}}"><img src="{{image}} /></a>{{model.customTagClose}}',
答案 1 :(得分:1)
除了史蒂文的回答,您可以使用此修复程序:
您可以使用此修复程序:
if (count % 4 === 0 || count === 0) {
image.customTagOpen = '<div>';
image.customTagClose = '';
} else if (count + 1 % 4 === 0) {
image.customTagOpen = '';
image.customTagClose = '</div>';
} else {
image.customTagOpen = '';
image.customTagClose = '';
}
干杯,
答案 2 :(得分:0)
对于div行每4个计数使用此:
filter: function(image) {
count += 1;
if (count == 1 || (count - 1) % 4 == 0 ) {
image.customTagOpen = '<div class="row">';
image.customTagClose = '';
} else if (count % 4 == 0) {
image.customTagOpen = '';
image.customTagClose = '</div>';
} else {
image.customTagOpen = '';
image.customTagClose = '';
}
return true;
},
或使用
filter: function(image) {
if (count == 0 || count % 4 == 0 ) {
image.customTagOpen = '<div class="row">';
image.customTagClose = '';
} else if ((count + 1) % 4 == 0) {
image.customTagOpen = '';
image.customTagClose = '</div>';
} else {
image.customTagOpen = '';
image.customTagClose = '';
}
count += 1;
return true;
},