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Solutions on the web

via GitHub by shizhy1986
, 1 year ago
via GitHub by yanunon
, 1 year ago
For input string: "0x20000"
via GitHub by marevol
, 9 months ago
For input string: ""
java.lang.NumberFormatException: For input string: "0x20000" at java.lang.NumberFormatException.forInputString( at java.lang.Integer.parseInt( at java.lang.Integer.valueOf( at net.dongliu.apk.parser.struct.xml.Attributes.getInt( at net.dongliu.apk.parser.parser.ApkMetaTranslator.onStartTag( at net.dongliu.apk.parser.parser.CompositeXmlStreamer.onStartTag( at net.dongliu.apk.parser.parser.BinaryXmlParser.readXmlNodeStartTag( at net.dongliu.apk.parser.parser.BinaryXmlParser.parse( at net.dongliu.apk.parser.AbstractApkParser.transBinaryXml( at net.dongliu.apk.parser.AbstractApkParser.parseManifestXml( at net.dongliu.apk.parser.AbstractApkParser.parseApkMeta( at net.dongliu.apk.parser.AbstractApkParser.getApkMeta( at