我想更改“car-image”类中所有图片的src。
但我不需要改变整个网址。我只想更改一个字符。
我想编辑这个 -
<div class="car-image">
<img src="/cars/3_large_1.png">
</div>
对此 -
<div class="car-image">
<img src="/cars/3_large_2.png">
</div>
这种格式在本课程的所有图像中都很常见。 我试过这样的事情 -
var allsrc = document.getElementsByClassName('car-image');
allsrc[0].src="/cars/3_large_2.png";
这不起作用。 我怎么能在javascript中做到这一点？
答案 0 :(得分：3)
你正在设置错误节点的src allsrc返回你的div而不是图像。
试试这个
var new_json = {{"nodes": [{ "id": 124587, "name": "paper1", "citation": 5, "group": 1 },
{ "id": 178456, "name": "paper2", "citation": 8, "group": 2 }],
"links": [{ "source": 124587, "target": 178456, "name": "A-B-1", "value": 8 }]};
function load_graph(text) {
var color = d3.scale.category20();
var data1 = {
"nodes": [
{ "id": 0, "name": "paper1", "citation": 5, "group": 1 },
{ "id": 1, "name": "paper2", "citation": 8, "group": 2 },
{ "id": 2, "name": "paper3", "citation": 12, "group": 3 },
{ "id": 3, "name": "paper4", "citation": 25, "group": 4 },
{ "id": 4, "name": "paper5", "citation": 15, "group": 5 },
{ "id": 5, "name": "paper6", "citation": 5, "group": 1 },
{ "id": 6, "name": "paper7", "citation": 8, "group": 2 },
{ "id": 7, "name": "paper8", "citation": 12, "group": 3 },
{ "id": 8, "name": "paper9", "citation": 25, "group": 4 },
{ "id": 9, "name": "paper10", "citation": 15, "group": 5 }
],
"links": [
{ "source": 0, "target": 1, "name": "A-B-1", "value": 8 },
{ "source": 0, "target": 1, "name": "A-B-2", "value": 24 },
{ "source": 0, "target": 2, "name": "A-C-1", "value": 12 },
{ "source": 0, "target": 2, "name": "A-C-3", "value": 44 },
{ "source": 2, "target": 3, "name": "A-D-1", "value": 11 },
{ "source": 2, "target": 3, "name": "A-D-2", "value": 35 },
{ "source": 2, "target": 4, "name": "A-E-1", "value": 16 },
{ "source": 2, "target": 4, "name": "A-E-5", "value": 30 },
{ "source": 4, "target": 5, "name": "A-B-1", "value": 8 },
{ "source": 4, "target": 5, "name": "A-B-2", "value": 24 },
{ "source": 5, "target": 6, "name": "A-C-1", "value": 12 },
{ "source": 5, "target": 6, "name": "A-C-3", "value": 44 },
{ "source": 5, "target": 7, "name": "A-D-1", "value": 11 },
{ "source": 5, "target": 7, "name": "A-D-2", "value": 35 },
{ "source": 7, "target": 8, "name": "A-E-1", "value": 16 },
{ "source": 7, "target": 8, "name": "A-E-5", "value": 30 },
{ "source": 8, "target": 3, "name": "A-C-1", "value": 12 },
{ "source": 8, "target": 3, "name": "A-C-3", "value": 44 },
{ "source": 8, "target": 9, "name": "A-D-1", "value": 11 },
{ "source": 8, "target": 9, "name": "A-D-2", "value": 35 }
]
};
// used to store the number of links between two nodes.
// mLinkNum[data.links[i].source + "," + data.links[i].target] = data.links[i].linkindex;
var mLinkNum = {};
// sort links first
sortLinks();
// set up linkIndex and linkNumer, because it may possible multiple links share the same source and target node
setLinkIndexAndNum();
var w = 960,
h = 500;
var force = d3.layout.force()
.nodes(d3.values(data1.nodes))
.links(data1.links)
.size([w, h])
.linkDistance(200)
.charge(-300)
.on("tick", tick)
.start();
var svg = d3.select(".graphContainer").append("svg:svg")
.attr("width", w)
.attr("height", h);
var path = svg.append("svg:g")
.selectAll("line")
.data(force.links())
.enter().append("svg:path")
.attr("class", "link")
.style("stroke-width", function (d) { return Math.sqrt(d.value); });
var circle = svg.append("svg:g")
.selectAll("circle")
.data(force.nodes())
.enter().append("svg:circle")
.attr("r", function (d) { return (d.citation); })
.style("fill", function (d) { return color(d.group); })
.call(force.drag);
var text = svg.append("svg:g")
.selectAll("g")
.data(force.nodes())
.enter().append("svg:g");
// A copy of the text with a thick white stroke for legibility.
text.append("svg:text")
.attr("x", 8)
.attr("y", ".31em")
.attr("class", "shadow")
.text(function (d) { return d.name; });
text.append("svg:text")
.attr("x", 8)
.attr("y", ".31em")
.text(function (d) { return d.name; });
// Use elliptical arc path segments to doubly-encode directionality.
function tick() {
path.attr("d", function (d) {
var dx = d.target.x - d.source.x,
dy = d.target.y - d.source.y,
dr = Math.sqrt(dx * dx + dy * dy);
// get the total link numbers between source and target node
var lTotalLinkNum = mLinkNum[d.source.id + "," + d.target.id] || mLinkNum[d.target.id + "," + d.source.id];
if (lTotalLinkNum > 1) {
// if there are multiple links between these two nodes, we need generate different dr for each path
dr = dr / (1 + (1 / lTotalLinkNum) * (d.linkindex - 1));
}
// generate svg path
return "M" + d.source.x + "," + d.source.y +
"A" + dr + "," + dr + " 0 0 1," + d.target.x + "," + d.target.y +
"A" + dr + "," + dr + " 0 0 0," + d.source.x + "," + d.source.y;
});
// Add tooltip to the connection path
path.append("svg:title")
.text(function (d, i) { return d.name; });
circle.attr("transform", function (d) {
return "translate(" + d.x + "," + d.y + ")";
});
text.attr("transform", function (d) {
return "translate(" + d.x + "," + d.y + ")";
});
}
// sort the links by source, then target
function sortLinks() {
data.links.sort(function (a, b) {
if (a.source > b.source) {
return 1;
}
else if (a.source < b.source) {
return -1;
}
else {
if (a.target > b.target) {
return 1;
}
if (a.target < b.target) {
return -1;
}
else {
return 0;
}
}
});
}
//any links with duplicate source and target get an incremented 'linknum'
function setLinkIndexAndNum() {
for (var i = 0; i < data.links.length; i++) {
if (i != 0 &&
data.links[i].source == data.links[i - 1].source &&
data.links[i].target == data.links[i - 1].target) {
data.links[i].linkindex = data.links[i - 1].linkindex + 1;
}
else {
data.links[i].linkindex = 1;
}
// save the total number of links between two nodes
if (mLinkNum[data.links[i].target + "," + data.links[i].source] !== undefined) {
mLinkNum[data.links[i].target + "," + data.links[i].source] = data.links[i].linkindex;
}
else {
mLinkNum[data.links[i].source + "," + data.links[i].target] = data.links[i].linkindex;
}
}
}
}
</script>
答案 1 :(得分：1)
我想要更改所有图像的src ＆＃39;汽车图像＆＃39;使用javascript的课程。
您可以为所有<img>类更改src car-image，如下所示：
var all = document.getElementsByClassName('car-image');
for(var i = 0; i < all.length; i++){
var image = document.getElementsByClassName('car-image')[i].getElementsByTagName('img');
image[0].setAttribute("src", "/cars/3_large_2.png");
}<div class="car-image">
<img src="/cars/3_large_1.png">
</div>
<div class="car-image">
<img src="/cars/5_large_1.png">
</div>
<div class="car-image">
<img src="/cars/7_large_1.png">
</div>
<div class="car-image">
<img src="/cars/9_large_1.png">
</div>
（检查元素并查看新的src＆＃39;
答案 2 :(得分：1)
更优雅的解决方案是使用带有正则表达式的替换功能。如果你知道图像src模式和类似的更改适用于所有图像src，你可以构建一个正则表达式。在这种情况下，不是逐个更改每个图像src，而是可以迭代包含car-image类的元素，找出第一个childNode并更改src attr。
// find all elements that contains class car-image
var carImgDivs = document.getElementsByClassName('car-image');
// iterate over carImgDivs and execute an imediate function to just pass the
// childNode1 that is the image. Use replace function with regex to find out the
// changed image src value and set the changed src value to childNode1
for(var i = 0; i < carImgDivs.length; i++) (function(childNode1) {
if(childNode1) {
var replacedSrc = childNode1.getAttribute('src').replace(/(_)(\d)/, "$12");
childNode1.setAttribute("src", replacedSrc);
}
})(carImgDivs[i].childNodes[1]);
对于像/cars/3_large_1.png这样的图像src，正则表达式（_）（\ d）匹配数字后面的下划线并捕获两者。替换字符串$1中的"$12"表示保持第一个捕获组（下划线）不变，而2表示用2替换第二个捕获组（一个数字） 。基本上，正则表达式与上面的图像src中的_1匹配。 _是第一个捕获组，1是第二个捕获组。因此，最后，图像src变为/cars/3_large_2.png
答案 3 :(得分：0)
使用（jQuery解决方案）：$( "img:nth-child(1)" ).attr('src', <new_name>);
nth-child(i)表示i图像。
示例：
$(".car_image img:nth-child(1)").attr('src', '/cars/3_large_2.png');
要更改所有图片，只需删除:nth-child()
答案 4 :(得分：0)
如果您include jquery in your page可以
$(".car-image img").attr("src", "/cars/3_large_2.png");
答案 5 :(得分：0)
这个怎么样？
var x = document.getElementsByClassName('car-image')[0];
var img = x.getElementsByTagName('img')[0];
img.src = "/cars/3_large_2.png";