How to escape % in String.Format?

Question

I am storing a SQL query in my strings.xml file and I want to use String.Format to build the final string in code. The SELECT statement uses a like, something like this:

SELECT Field1, Field2 FROM mytable WHERE Field1 LIKE '%something%'

In order to format that I replace 'something' with %1$s so it becomes:

SELECT Field1, Field2 FROM mytable WHERE Field1 LIKE \'%%1$s%\'

I escape the single quotes with the backslash. However I am not able to escape the % sign.

How can I include a like statement in my strings.xml file?

They'd be injecting into their own database, no concern here ;) — Matthew
– Matthew, Commented Feb 16, 2011 at 3:06
Well, even if it is Your own database, it is possible to accidentally write queries that do bad things. Or just write queries that do not compile. Preparing queries is a good habit to get into. — Rauni Lillemets
– Rauni Lillemets, Commented Nov 28, 2014 at 9:16
Although it's slower than String.format() you might consider using MessageFormat() instead. — ccpizza
– ccpizza, Commented May 17, 2018 at 9:53

limc · Accepted Answer · 2012-07-11 13:50:32Z

1218

To escape %, you will need to double it up: %%.

edited Jul 11, 2012 at 13:50

user142162

answered Feb 16, 2011 at 2:40

limc

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Cavaleiro · Accepted Answer · 2015-07-21 17:39:41Z

32

To complement the previous stated solution, use:

str = str.replace("%", "%%");

answered Jul 21, 2015 at 17:39

Cavaleiro

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2 Comments

Toilal Over a year ago

This will replace % that are already doubled. Please read the other answer.

David Balažic Over a year ago

@Toilal Why would anyone escape something that is already escaped? And if he would, why not actually do it? Maybe he intended to have two % signs, so the correct escaped form would be '%%%%'

Toilal · Accepted Answer · 2016-07-27 06:43:49Z

5

This is a stronger regex replace that won't replace %% that are already doubled in the input.

str = str.replaceAll("(?:[^%]|\\A)%(?:[^%]|\\z)", "%%");

edited Jul 27, 2016 at 6:43

answered Jul 27, 2016 at 6:38

Toilal

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1 Comment

Damian Helme Over a year ago

seems to be an error with the regex: "76%".replaceAll("(?:[^%]|\\A)%(?:[^%]|\\z)", "%%") gives "7%%"

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