fix 未完全解决的代码生成字典问题

master
Nguyendream 5 years ago committed by Gitee
parent 28fa991fc2
commit 1031ba44b7

@ -105,7 +105,7 @@
<span>{{ parseTime(scope.row.${javaField}, '{y}-{m}-{d}') }}</span> <span>{{ parseTime(scope.row.${javaField}, '{y}-{m}-{d}') }}</span>
</template> </template>
</el-table-column> </el-table-column>
#elseif($column.list && "" != $column.dictType) #elseif($column.list && $column.dictType && "" != $column.dictType)
<el-table-column label="${comment}" align="center" prop="${javaField}" :formatter="${javaField}Format" /> <el-table-column label="${comment}" align="center" prop="${javaField}" :formatter="${javaField}Format" />
#elseif($column.list && "" != $javaField) #elseif($column.list && "" != $javaField)
#if(${foreach.index} == 1) #if(${foreach.index} == 1)

@ -134,7 +134,7 @@
<span>{{ parseTime(scope.row.${javaField}, '{y}-{m}-{d}') }}</span> <span>{{ parseTime(scope.row.${javaField}, '{y}-{m}-{d}') }}</span>
</template> </template>
</el-table-column> </el-table-column>
#elseif($column.list && "" != $column.dictType) #elseif($column.list && $column.dictType && "" != $column.dictType)
<el-table-column label="${comment}" align="center" prop="${javaField}" :formatter="${javaField}Format" /> <el-table-column label="${comment}" align="center" prop="${javaField}" :formatter="${javaField}Format" />
#elseif($column.list && "" != $javaField) #elseif($column.list && "" != $javaField)
<el-table-column label="${comment}" align="center" prop="${javaField}" /> <el-table-column label="${comment}" align="center" prop="${javaField}" />

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