拿。 postgres 中 TEXT 中的特定字符串

Question

我有一个问题：下面是存储在 postgres table 列中的一条错误消息，从这个字符串中我只想提取字符串的一部分，在 Postgres 中可以这样做吗？

我想看odoo.exceptions.ValidationError: ('No MRP template was found for MO/10881!', None)' 只有这一部分。

通常所有以 odoo.exceptions.ValidationError 开头的文本：直到结尾

我该怎么做？有什么想法或建议吗？

'Traceback (most recent call last):
  File "/opt/src/addons_OCA/queue/queue_job/controllers/main.py", line 101, in runjob
    self._try_perform_job(env, job)
  File "/opt/src/addons_OCA/queue/queue_job/controllers/main.py", line 61, in _try_perform_job
    job.perform()
  File "/opt/src/addons_OCA/queue/queue_job/job.py", line 466, in perform
    self.result = self.func(*tuple(self.args), **self.kwargs)
  File "/opt/src/addons/costs/models/mrp_production.py", line 163, in trigger_calculate_all_costs
    self.calculate_all_costs()
  File "/opt/src/addons/sucosts/models/costline_mixin.py", line 284, in calculate_all_costs
    rec.generate_service_products()
  File "/opt/src/addons/mrp_product_templates/models/mrp_production.py", line 660, in generate_service_products
    MO=self.name)))
odoo.exceptions.ValidationError: ('No MRP template was found for MO/10881!', None)'

Answer 1

您可以使用 regexp_replace 函数搜索特定文本，然后 select 后面的文本。

The regexp_replace function provides substitution of new text for substrings that match POSIX regular expression patterns. It has the syntax regexp_replace(source, pattern, replacement [, flags ]). The source string is returned unchanged if there is no match to the pattern. ...

在这种情况下，您似乎需要 ValidationError: 之后的文本。类似于：（参见 demo）

select regexp_replace (message, '.*ValidationError:(.*)','') 
  from test;

拿。 postgres 中 TEXT 中的特定字符串

Take. a particular string from a TEXT in postgres

postgresql