oracle分割字符串后以单列多行展示
原始的sql: select substr(field1, instr(field1, |, 1, rownum) + 1, instr(field1, |, 1, rownum + 1) - instr(field1, |, 1, rownum) - 1) as field2 from (select | || a|bbb|cccc|ddddd|ee|d|a || | as field1 from dual)connect by instr(field1, |,
原始的sql:
select substr(field1, instr(field1, '|', 1, rownum) + 1, instr(field1, '|', 1, rownum + 1) - instr(field1, '|', 1, rownum) - 1) as field2 from (select '|' || 'a|bbb|cccc|ddddd|ee|d|a' || '|' as field1 from dual) connect by instr(field1, '|', 2, rownum) > 0;
查询结果如下:
field2
a
bbb
cccc
ddddd
ee
d
a
connect by :递归,即查询继续的条件
instr(field1, '|', 2, rownum) 字符串从第2个位置开始,即从a开始,‘|’分隔符从第一次到第八次出现的位置依次是:3、5、7、9、11、13、15、0,所以会查询七次,也就是会有七行。
substr函数的第二个参数为instr(field1, '|', 1, rownum) + 1,表示‘|’从第一个位置开始匹配,第一到第七次匹配的位置加上一,即2、4、6、8、10、12、14
substr函数的第三个参数仔细看一下,恒等于一。
也就是说,按‘|’分割后,第一次取第二部分,第二次取第四部分,依次类推。
取的值分别为:a、bbb、cccc、ddddd、ee、d、a
因为rownum是递增的,所以会变成七行数据。
分析完毕。

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