Bug #41429 REPLACE command could not be formatted

Submitted: 12 Dec 2008 12:42
Reporter: Mark Leith Email Updates:
Status: Verified Impact on me:
None 
Category:MySQL Enterprise Monitor: Web Severity:S3 (Non-critical)
Version:2.0.0.7122 OS:Any
Assigned to: Assigned Account CPU Architecture:Any
Tags: formatter, sqlparser

[12 Dec 2008 12:42] Mark Leith 
Description:
The following statement could not be formatted:

REPLACE INTO support_issue_bug ( issue_id , bug_id , bug_category , bug_status , bug_reported_date , is_currently_associated ) SELECT si . issue_id , bdb . id , bdb . bug_type , bdb . status , bdb . ts1 , ? FROM support_issue si JOIN bugs . bugdb bdb ON ( bdb . affectedissues IS NOT NULL AND bdb . affectedissues != ? AND FIND_IN_SET( REPLACE( bdb . affectedissues , ? , ? ) , si . issue_id ) )

The logs report:

com.mysql.sqlparser.MySQLParser	Warning	Dec 12, 2008 12:37:45 PM	line 1:0 no viable alternative at input [@0,0:6='REPLACE',<148>,1:0]
com.mysql.sqlparser.MySQLParser	Warning	Dec 12, 2008 12:37:45 PM	line 1:0 no viable alternative at input [@0,0:6='REPLACE',<148>,1:0]

How to repeat:
N/A

Suggested fix:
N/A

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