我正在尝试从
获取Call:
lm(formula = y ~ x1 + x2 + x3 + z1 + z2 + z3, data = df)
Residuals:
Min 1Q Median 3Q Max
-2.76472 -0.56958 -0.02673 0.50188 2.61362
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.18092 0.09966 1.815 0.0727 .
x1 0.12282 0.10231 1.201 0.2330
x2 -0.22411 0.10781 -2.079 0.0404 *
x3 -0.01096 0.09554 -0.115 0.9090
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.9596 on 93 degrees of freedom
Multiple R-squared: 0.07717, Adjusted R-squared: 0.01763
F-statistic: 1.296 on 6 and 93 DF, p-value: 0.2667
的子字符串
The access token provided is invalid我正在做的是
Access to the requested resource path is unauthorized: v1/streaming/video/558489e46b66d1023309e1a1 [The access token provided is invalid]
但 NSError *err = nil;
NSRegularExpression *regex = [NSRegularExpression regularExpressionWithPattern:@"[(.*)]" options:0 error:&err];
NSString *message = nil;
if (regex) {
NSTextCheckingResult *result = [regex firstMatchInString:(NSString*)error.userInfo[@"NSLocalizedDescription"]
options:0
range:NSMakeRange(0, ((NSString*)error.userInfo[@"NSLocalizedDescription"]).length)];
if ( result ) {
NSRange range = [result rangeAtIndex:1];
message = [((NSString*)error.userInfo[@"NSLocalizedDescription"]) substringWithRange:range];
}
}
为零。我的想法是在message之间找到任何单词。我认为我的常规有问题，因为我使用[]。
有没有人对这个问题有任何暗示。
答案 0 :(得分：0)
ICU用户指南：Regular Expressions
正则表达式：(?<= \\[)[^\\]]+(?=\\])
(?<= \\[)后视断言[必须逃脱的
[^\\]]+一个或多个必须转义的字符
(?=\\])前瞻性断言，必须转义
NSString *string = @"Access to the requested resource path is unauthorized: v1/streaming/video/558489e46b66d1023309e1a1 [The access token provided is invalid]";
NSString *regex = @"(?<= \\[)[^\\]]+(?=\\])";
NSRange range = [string rangeOfString:regex options:NSRegularExpressionSearch];
NSLog(@"range: %@", NSStringFromRange(range));
NSLog(@"found: %@", [string substringWithRange:range]);
输出：
范围：{100,36}
found：提供的访问令牌无效