How to pass the content calculated by the function into a variable.
<!DOCTYPE html>
<html lang="zh">
<head>
<title>照片爬取</title>
<meta charset="UTF-8">
</head>
<body>
<script>
var a = 0;
function photo() {
while (a < 10) {
a = a + 1;
var url = "http://work.ynzs.cn/ZSGL/servletphoto?path=kszp/22240205/300y00" + a + ".jpg";
document.write(url + ",");
}
while (a < 100) {
a = a + 1;
var url = "http://work.ynzs.cn/ZSGL/servletphoto?path=kszp/22240205/300y0" + a + ".jpg";
document.write(url + ",");
}
}
</script>
<input name="button" type="button" onClick="photo()" value="NEXT" />
</body>
</html>Like the code, use this calculation to output the image address I want, and then use the following code to display the image on the page
<script>
var img = “图片地址”;
var body = document.querySelector('body');
var imgArr = img.split(",");
var tem, j = 0;
document.querySelector('#showall').addEventListener('click', showa);
function showa() {
for (j; j < imgArr.length; j++) {
tem = document.createElement('img');
tem.src = imgArr[j];
body.appendChild(tem);
}
}
</script>
<input name="button" type="button" onClick="showa()" value="show" />But how to pass the content calculated by the first paragraph of code to the second paragraph? ? ?
I am a JS beginner, please give me some advice
phpcn_u15822017-06-26 10:57:07
I think it’s okay to encapsulate the second code into a function that receives the image address as a parameter
高洛峰2017-06-26 10:57:07
Judging from the meaning of the title, you return the calculated value inside the function, and when running this function, just use a variable to receive it.
