本文介绍了试图利用JSZip打开然后解析.zip中的特定文件的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
试图使用在.zip中循环浏览文件,寻找一个(这里, test.txt
)我想要解析的内容。
Have been trying to use the JSZip library to cycle through files in a .zip, looking for one (here, test.txt
) that I want to parse for content.
的 [推荐查看源代码] JSZip提供:
Have attempted to do a modification of the sample [recommend viewing source on that] that JSZip provides:
<!DOCTYPE HTML>
<html>
<head>
<link href="https://netdna.bootstrapcdn.com/twitter-bootstrap/2.3.1/css/bootstrap-combined.min.css" rel="stylesheet">
<link href="https://netdna.bootstrapcdn.com/twitter-bootstrap/2.3.1/css/bootstrap-combined.min.css" rel="stylesheet">
</head>
<body>
<div class = "container">
<div class = "hero-unit">
<input type="file" class="span7" id="input" name="file" multiple /> <!-- redo this in a bootstrappy way-->
<br>
<output id="output"></output>
</div>
</div>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.0/jquery.min.js"></script>
<script src="/js/jszip-load.js"></script>
<script src="/js/jszip.js"></script>
<script src="/js/jszip-inflate.js"></script>
<script>
if (window.File && window.FileReader && window.FileList && window.Blob) {
// Great success! All the File APIs are supported.
} else {
alert('The File APIs are not fully supported in this browser.');
}
function handleFileSelect(evt) {
var files = evt.target.files; // FileList object
// files is a FileList of File objects. List some properties.
var output = [];
for (var i = 0, f; f = files[i]; i++) {
if (f.type !== "application/zip") {
document.getElementById('output').innerHTML = "<p class='text-error'>" + f.name + " isn't a zip file.</div>";
continue;
}
var reader = new FileReader();
reader.onload = (function(theFile) {
return function(e) {
var zip = new JSZip(e.target.result)
$.each(zip.files, function (index, zipEntry) {
if (zipEntry.name == "test.txt"){
var text = zipEntry.asText();
var lines = text.split(/[\r\n]+/g); // tolerate both Windows and Unix linebreaks
for(var i = 0; i < lines.length; i++) {
if (lines[i].length > 240){
output.push('<li>' + lines[i] + '<br>');
}
}
document.getElementById('output').innerHTML = '<h2>Paths with more than 240 characters:</h2> <br><ol>' + output.join('') + '</ol>';
else{
alert("file not found!")
}
}
});
}
})(f);
}
}
document.getElementById('input').addEventListener('change', handleFileSelect, false);
</script>
</body>
</html>
由于某种原因,我确信必须使用我使用闭包的方式,它实际上不是解析有问题的.zip文件。任何想法我可能在这里做错了?
For some reason, I'm sure having to do with the way that I am using the closure, it is not actually parsing the .zip files in question. Any ideas what I might be doing wrong here?
推荐答案
我使用这个代码,并能够得到所有的文件数据。内容变量具有文件内容:
I use this code and am able to get all file data. content variable has file content:
function loadSettingsFile(evt) {
var files = evt.target.files;
for (var i = 0, f; f = files[i]; i++) {
var reader = new FileReader();
// Closure to capture the file information.
reader.onload = (function(theFile) {
return function(e) {
try {
var zip = new JSZip(e.target.result);
$.each(zip.files, function (index, zipEntry) {
var content = zipEntry.asText();
alert(content);
});
} catch(e) {
alert(e)
}
}
})(f);
// read the file !
// readAsArrayBuffer and readAsBinaryString both produce valid content for JSZip.
reader.readAsArrayBuffer(f);
// reader.readAsBinaryString(f);
}
}
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