Question

我有2个表，第一个表#include <stdio.h> #include <string.h> int main(int argc, const char *argv[]) { const char *srcStr = "<AAA>hello world</AAA> this is a text <AAA>this is another text</AAA>"; const char *firstDelim = "<AAA>"; const char *lastDelim = "</AAA>"; char extStr[128] = ""; int srcStrLen, firstDelimLen, lastDelimLen; int i, catIdx = 0; srcStrLen = strlen(srcStr); firstDelimLen = strlen(firstDelim); lastDelimLen = strlen(lastDelim); for (i=0; i<srcStrLen; i++){ if (strncmp(&srcStr[i], firstDelim, firstDelimLen) == 0) { i += firstDelimLen; catIdx = i; } else if (strncmp(&srcStr[i], lastDelim, lastDelimLen) == 0) { strncat(extStr, &srcStr[catIdx], i - catIdx); i += lastDelimLen; if (i != srcStrLen){ strcat(extStr, " "); } } } printf("%s\n", extStr); return 0; }数据。第二个表or_f_table

or_table

我想做这样的结果： -

or_f_table
f_id   | f_o_id  | f_u_id
1      |    19   |    1
2      |    5    |    2
3      |    19   |    2

or_table
o_id  | o_name
  4   |  test1
  5   |  test2
  19  |  oops2
  20  |  oops3


SELECT o.o_name,
IF ((SELECT count(*)  FROM or_f_table as f 
WHERE f.f_u_id = 1 ),'Yes','No') as follow_status
FROM or_table as o  
WHERE o.o_name LIKE '%oop%'

我收到了结果

o_name   |  follow_status
oops2    |  Yes
oops3    |  No

为什么它不起作用？我应该如何纠正呢

Answer 1

对于where条件，总会有一个大于0的值。这就是为什么它不起作用。

尝试此操作以获得指定的结果

    SELECT o.o_name,
     IF ((SELECT count(*)  FROM or_f_table as f 
     WHERE f.f_o_id = o.o_id ),'Yes','No') as follow_status
     FROM or_table as o  
     WHERE o.o_name LIKE '%oop%'

如果是或否状态mysql查询

1 个答案: