Changeset 34456 in webkit
- Timestamp:
- Jun 8, 2008 1:57:48 PM (16 years ago)
- Location:
- trunk/WebCore
- Files:
-
- 1 added
- 2 edited
Legend:
- Unmodified
- Added
- Removed
-
trunk/WebCore/ChangeLog
r34455 r34456 1 2008-06-08 Adam Roben <aroben@apple.com> 2 3 Fix Bug 18837: Database panel fails to display tables if any value is 4 NULL 5 6 <https://bugs.webkit.org/show_bug.cgi?id=18837> 7 8 Reviewed by Darin Adler. 9 10 Test: manual-tests/inspector/display-sql-null.html 11 12 * manual-tests/inspector/display-sql-null.html: Added. 13 * page/inspector/DatabasesPanel.js: 14 (WebInspector.DatabasesPanel.prototype.dataGridForResult): Convert all 15 objects to strings before operating on them. 16 1 17 2008-06-08 Adam Roben <aroben@apple.com> 2 18 -
trunk/WebCore/page/inspector/DatabasesPanel.js
r34072 r34456 186 186 var row = rows.item(i); 187 187 for (var columnIdentifier in row) { 188 var text = row[columnIdentifier]; 188 // FIXME: (Bug 19439) We should specially format SQL NULL here 189 // (which is represented by JavaScript null here, and turned 190 // into the string "null" by the String() function). 191 var text = String(row[columnIdentifier]); 189 192 data[columnIdentifier] = text; 190 193 if (text.length > columns[columnIdentifier].width)
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