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中引用它的行。所以这是一个将被重构的函数的怪物,但现在我只是试图让它工作。谷歌地图javascript:如果数组[i]是未定义的,忽略在infowindow.setContent
问题是,打开标记infowindow失败时,它包含未定义的谷歌地方变量。我希望能够忽略infowindow.setContent中引用未定义变量的行,以便无论Google地方信息可用,infowindow都会打开。我可以单独检查每个人,但希望有人能找到更有说服力的解决方案。
我的代码:
//passed a google place object
function formatInfoWindow(place) {
var name = place.name,
open = place.opening_hours.open_now,
address = place.vicinity,
phoneI = place.international_phone_number,
phone = place.formatted_phone_number
rating = place.rating,
reviewAuth = place.reviews[0].author_name,
reviewText = place.reviews[0].text,
reviewRate = place.reviews[0].rating,
website = place.website,
img = place.photos[0],
photo = img.getUrl(),
info = this.info;
//creates an array of all the place variables
var infoWindowArray = [name, open, address,phoneI, phone, rating, reviewAuth, reviewText, reviewRate, website, img, photo, info];
//loop through array to find undefined variables
for (var i = 0; i < infoWindowArray.length; i++) {
if (typeof infoWindowArray[i] === 'undefined') {
//somehow ignore those variables in setContent block
}
}
this.infowindow.setContent(
'<div class="infowindow"><strong><h1>' + name + '</h1></strong>' +
'<address>' + address + '</address>'+
'<p>Open Now: ' + open + ' Rating: ' + rating + ' ' + '<a href="' + website + '">Website</a></p>'+
'<p><a href="tel:' + phoneI + '">'+phone+'</a></p>'+
'<h3><b>Notes:</b></h3>'+
'<p>' + info + '</p>'+
'<h4>' + reviewAuth + ' Rating: '+reviewRate+'</h4>'+
'<p>' + reviewText + '</p>'+
'<img src='+ photo + ' ></img>'+
'</div>'
);
this.infowindow.open(map, this);
}
谢谢!我试图使用模板字符串,但解决方案不断变得复杂,所以我放弃了它。太棒了。 –
你能解释一下这个函数吗(match,prop){ return infos [prop] || “”; }); –
更多信息在这里:https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/replace – Diode