Develop Reference

Csv: search for String and replace with another string

I have a .csv file that contains:
scenario, custom, master_data
1, ${CUSTOM}, A_1
I have a string:
a, b, c
and I want to replace 'custom' with 'a, b, c'. How can I do that and save to the existing .csv file?

Probably the easiest way is to read in one file and output to another file as you go, modifying it on a per-line basis
You could try something with tokenizers, this may not be completely correct for your output/input, but you can adapt it to your CSV file formatting
BufferedReader reader = new BufferedReader(new FileReader("input.csv"));
BufferedWriter writer = new BufferedWriter(new FileWriter("output.csv"));
String custom = "custom";
String replace = "a, b, c";
for(String line = reader.readLine(); line != null; line = reader.readLine())
{
String output = "";
StringTokenizer tokenizer = new StringTokenizer(line, ",");
for(String token = tokenizer.nextToken(); tokenizer.hasMoreTokens(); token = tokenizer.nextToken())
if(token.equals(custom)
output = "," + replace;
else
output = "," + token;
}
readInventory.close();
If this is for a one off thing, it also has the benefit of not having to research regular expressions (which are quite powerful and useful, good to know, but maybe for a later date?)

Have a look at Can you recommend a Java library for reading (and possibly writing) CSV files?
And once the values have been read, search for strings / value that start with ${ and end with }. Use Java Regular Expressions like \$\{(\w)\}. Then use some map for looking up the found key, and the related value. Java Properties would be a good candidate.
Then write a new csv file.

Since your replacement string is quite unique you can do it quickly without complicated parsing by just reading your file into a buffer, and then converting that buffer into a string. Replace all occurrences of the text you wish to replace with your target text. Then convert the string to a buffer and write that back to the file...
Pattern.quote is required because your string is a regular expression. If you don't quote it you may run into unexpected results.
Also it's generally not smart to overwrite your source file. Best is to create a new file then delete the old and rename the new to the old. Any error halfway will then not delete all your data.
final Path yourPath = Paths.get("Your path");
byte[] buff = Files.readAllBytes(yourPath);
String s = new String(buff, Charset.defaultCharset());
s = s.replaceAll(Pattern.quote("${CUSTOM}"), "a, b, c");
Files.write(yourPath, s.getBytes());

Java Replacing Help Needed

Hey guy's so am trying to replace all characters and numbers to get the /hello/what/ only without the REMOVEThis4.PNG i don't want to use string.replace("REMOVEThis4.PNG", ""); cause i wanna use it on other strings not only that
Any help is great my code
String sFile = "/hello/what/REMOVEThis4.PNG";
if (sFile.contains("/")){
String Replaced = sFile.replaceAll("(?s)", "");
System.out.println(Replaced);
}
I want the the output to be
/hello/what/
Only thanks alot!

If you are trying to parse a path, I recommend to find the last index of /, and get the substring to this index plus one. So
string = string.substring(0, string.lastIndexOf("/") + 1);

No need to use regular expressions in your case:
String sFile = "/hello/what/REMOVEThis4.PNG";
// TODO check actual last index of "/" against -1
System.out.println(sFile.substring(0, sFile.lastIndexOf("/") + 1));
Output
/hello/what/
Note
In case you are dealing with actual files, you can probably spare yourself the String manipulation and use File.getParent() instead:
File file = new File("/hello/what/REMOVEThis4.PNG");
System.out.println(file.getParent());
Output (may change depending on your system)
\hello\what

Use Java's File API:
String example = "/hello/what/REMOVEThis4.PNG";
File file = new File(example);
System.out.println(example);
String absolutePath = file.getAbsolutePath();
String filePath = absolutePath.substring(0, absolutePath.lastIndexOf(File.separator));
System.out.println(filePath);

concatenate a string and a null character

I am completely new to java, and I am trying to join to string together with a null character between them.
String header = "text 19"
String content = "this is the content"
String wholething = // this is supposed to be "text 19\0this is the content"
End-goal here is to take sha-1 sum of the wholething and write the wholething to a file. I have tried some google searches, but couldn't figure out how to do it.

String wholething = header + '\0' + content;
Other ways to concatenate strings (in no particular order), each with their own particular uses:
String.concat() - Essentially the same as the + operator.
String.format() - A flexible way to build a string.
StringBuilder - Efficient way to concatenate many strings.
StringBuffer - Similar to StringBuilder, but thread-safe.
StringWriter - Extends Writer, compatible with stream IO functions.
Probably a couple more here and there.

String wholeThing = header + '\0' + content;

String wholething = header + "\0" + content;

Getting the last token String

I have a application that needs to read a String buffer that is semi-colon ';' delimited.
String buff = foo.getBuff(); // returns the buffer
However, the buffer can get pretty large and I just need to get the last String token and then flush the temporary variable String buff so my app won't accumulate much memory.
Update:
I tried this code:
String lastToken = buff.substring(buff.lastIndexOf(";") + 1);
However, I am not getting result with this code above, compared to this:
List<String> slist = Arrays.asList(buff.split(";"));
String lastToken = slist.get(slist.size() - 1);
However using List is very slow. My web app is almost not responding when processing this.

This will help you:
String lastToken = buff.substring(buff.lastIndexOf(";") + 1);

I don't know Java, but by googling 2 functions, I think something like this should work:
String buff = foo.getBuff(); // returns the buffer
String lastToken = buff.substring(buff.lastIndexOf(";")+1);

Check this:
String lastToken = foo.getBuff().substring(foo.getBuff().lastIndexOf(';')+1);
Edit: Dan gave the answer better and faster than me.

Java - Convert String to valid URI object

I am trying to get a java.net.URI object from a String. The string has some characters which will need to be replaced by their percentage escape sequences. But when I use URLEncoder to encode the String with UTF-8 encoding, even the / are replaced with their escape sequences.
How can I get a valid encoded URL from a String object?
http://www.google.com?q=a b gives http%3A%2F%2www.google.com... whereas I want the output to be http://www.google.com?q=a%20b
Can someone please tell me how to achieve this.
I am trying to do this in an Android app. So I have access to a limited number of libraries.

You might try: org.apache.commons.httpclient.util.URIUtil.encodeQuery in Apache commons-httpclient project
Like this (see URIUtil):
URIUtil.encodeQuery("http://www.google.com?q=a b")
will become:
http://www.google.com?q=a%20b
You can of course do it yourself, but URI parsing can get pretty messy...

Android has always had the Uri class as part of the SDK:
http://developer.android.com/reference/android/net/Uri.html
You can simply do something like:
String requestURL = String.format("http://www.example.com/?a=%s&b=%s", Uri.encode("foo bar"), Uri.encode("100% fubar'd"));

I'm going to add one suggestion here aimed at Android users. You can do this which avoids having to get any external libraries. Also, all the search/replace characters solutions suggested in some of the answers above are perilous and should be avoided.
Give this a try:
String urlStr = "http://abc.dev.domain.com/0007AC/ads/800x480 15sec h.264.mp4";
URL url = new URL(urlStr);
URI uri = new URI(url.getProtocol(), url.getUserInfo(), url.getHost(), url.getPort(), url.getPath(), url.getQuery(), url.getRef());
url = uri.toURL();
You can see that in this particular URL, I need to have those spaces encoded so that I can use it for a request.
This takes advantage of a couple features available to you in Android classes. First, the URL class can break a url into its proper components so there is no need for you to do any string search/replace work. Secondly, this approach takes advantage of the URI class feature of properly escaping components when you construct a URI via components rather than from a single string.
The beauty of this approach is that you can take any valid url string and have it work without needing any special knowledge of it yourself.

Even if this is an old post with an already accepted answer, I post my alternative answer because it works well for the present issue and it seems nobody mentioned this method.
With the java.net.URI library:
URI uri = URI.create(URLString);
And if you want a URL-formatted string corresponding to it:
String validURLString = uri.toASCIIString();
Unlike many other methods (e.g. java.net.URLEncoder) this one replaces only unsafe ASCII characters (like ç, é...).
In the above example, if URLString is the following String:
"http://www.domain.com/façon+word"
the resulting validURLString will be:
"http://www.domain.com/fa%C3%A7on+word"
which is a well-formatted URL.

If you don't like libraries, how about this?
Note that you should not use this function on the whole URL, instead you should use this on the components...e.g. just the "a b" component, as you build up the URL - otherwise the computer won't know what characters are supposed to have a special meaning and which ones are supposed to have a literal meaning.
/** Converts a string into something you can safely insert into a URL. */
public static String encodeURIcomponent(String s)
{
StringBuilder o = new StringBuilder();
for (char ch : s.toCharArray()) {
if (isUnsafe(ch)) {
o.append('%');
o.append(toHex(ch / 16));
o.append(toHex(ch % 16));
}
else o.append(ch);
}
return o.toString();
}
private static char toHex(int ch)
{
return (char)(ch < 10 ? '0' + ch : 'A' + ch - 10);
}
private static boolean isUnsafe(char ch)
{
if (ch > 128 || ch < 0)
return true;
return " %$&+,/:;=?#<>#%".indexOf(ch) >= 0;
}

You can use the multi-argument constructors of the URI class. From the URI javadoc:
The multi-argument constructors quote illegal characters as required by the components in which they appear. The percent character ('%') is always quoted by these constructors. Any other characters are preserved.
So if you use
URI uri = new URI("http", "www.google.com?q=a b");
Then you get http:www.google.com?q=a%20b which isn't quite right, but it's a little closer.
If you know that your string will not have URL fragments (e.g. http://example.com/page#anchor), then you can use the following code to get what you want:
String s = "http://www.google.com?q=a b";
String[] parts = s.split(":",2);
URI uri = new URI(parts[0], parts[1], null);
To be safe, you should scan the string for # characters, but this should get you started.

I had similar problems for one of my projects to create a URI object from a string. I couldn't find any clean solution either. Here's what I came up with :
public static URI encodeURL(String url) throws MalformedURLException, URISyntaxException
{
URI uriFormatted = null;
URL urlLink = new URL(url);
uriFormatted = new URI("http", urlLink.getHost(), urlLink.getPath(), urlLink.getQuery(), urlLink.getRef());
return uriFormatted;
}
You can use the following URI constructor instead to specify a port if needed:
URI uri = new URI(scheme, userInfo, host, port, path, query, fragment);

Well I tried using
String converted = URLDecoder.decode("toconvert","UTF-8");
I hope this is what you were actually looking for?

The java.net blog had a class the other day that might have done what you want (but it is down right now so I cannot check).
This code here could probably be modified to do what you want:
http://svn.apache.org/repos/asf/incubator/shindig/trunk/java/common/src/main/java/org/apache/shindig/common/uri/UriBuilder.java
Here is the one I was thinking of from java.net: https://urlencodedquerystring.dev.java.net/

Or perhaps you could use this class:
http://developer.android.com/reference/java/net/URLEncoder.html
Which is present in Android since API level 1.
Annoyingly however, it treats spaces specially (replacing them with + instead of %20). To get round this we simply use this fragment:
URLEncoder.encode(value, "UTF-8").replace("+", "%20");

I ended up using the httpclient-4.3.6:
import org.apache.http.client.utils.URIBuilder;
public static void main (String [] args) {
URIBuilder uri = new URIBuilder();
uri.setScheme("http")
.setHost("www.example.com")
.setPath("/somepage.php")
.setParameter("username", "Hello Günter")
.setParameter("p1", "parameter 1");
System.out.println(uri.toString());
}
Output will be:
http://www.example.com/somepage.php?username=Hello+G%C3%BCnter&p1=paramter+1