我有以下复杂的查询,希望在Laravel中使用。但直到现在我都没有成功。也许你们中的一个可以演示如何将其“翻译”到Laravel的查询生成器。
这里,Workbench中的原始查询运行良好:
select * from lemma l
inner join etymology e on l.id=e.lemma_id_fk
inner join gloss g on e.id = g.etymology_id
inner join wold_meanings w on g.gloss=w.meaning
where g.gloss like '%flower%' #paper, ocean, etc.
limit 100;比我试过的更好:
$results = DB::table('lemma')
->join('etymology', 'lemma.id', '=', 'etymology.lemma_id_fk')
->join('gloss', 'etymology.id', '=', 'gloss.etymology_id')
->join('wold_meanings', 'gloss.gloss', '=', 'wold_meanings.meaning')
->select(DB::raw('lemma.*'))
->where('gloss.gloss', 'like', '%flower%')
->get();还包括:
$results = DB::table('lemma')
->select(DB::raw("select * from lemma l
inner join etymology e on l.id=e.lemma_id_fk
inner join gloss g on e.id = g.etymology_id
inner join wold_meanings w on g.gloss=w.meaning
where g.gloss like '%flower%'
limit 1"));这也是:
$results = DB::raw("select * from lemma l
inner join etymology e on l.id=e.lemma_id_fk
inner join gloss g on e.id = g.etymology_id
inner join wold_meanings w on g.gloss=w.meaning
where g.gloss like '%flower%'
limit 100");但我绝对没有成功..。:-(
有人要给我指路吗?我用的是Laravel5.4。
==编辑#1 ==
在这里,应该显示结果的视图:
<table class="table table-bordered table-condensed table-hover table-responsive table-striped" id="table">
<thead>
<tr>
<th>id</th>
<th>lemma_id</th>
<th>headword</th>
<th>lemma</th>
<th>pos</th>
<th>gender</th>
<th>language</th>
<th>origin_family</th>
<th>origin</th>
<th>short_path</th>
<th>origin_path</th>
<th>etymology_text</th>
<th>first_use</th>
<th>lang</th>
<th>pageid</th>
<th>term</th>
<th>non_latin_script</th>
<th>lang_2</th>
<th>gloss</th>
<th>sequence</th>
<th>lemma_id_fk</th>
<th>derivatives_id</th>
<th>etymology_id</th>
<th>meaning</th>
<th>semantic_category</th>
<th>semantic_field</th>
<th>simplicity_score</th>
<th>age_score</th>
<th>borrowed_score</th>
<th>description</th>
<th>typical_context</th>
<th>representation</th>
<th>sub_code</th>
</tr>
</thead>
<tbody>
@foreach($results as $result)
<tr>
<td>{{$result->id}}</td>
<td>{{$result->lemma_id}}</td>
<td>{{$result->headword}}</td>
<td>{{$result->lemma}}</td>
<td>{{$result->pos}}</td>
<td>{{$result->gender}}</td>
<td>{{$result->language}}</td>
<td>{{$result->origin_family}}</td>
<td>{{$result->origin}}</td>
<td>{{$result->short_path}}</td>
<td>{{$result->origin_path}}</td>
<td>{{$result->etymology_text}}</td>
<td>{{$result->first_use}}</td>
<td>{{$result->lang}}</td>
<td>{{$result->pageid}}</td>
<td>{{$result->term}}</td>
<td>{{$result->non_latin_script}}</td>
<td>{{$result->lang_2}}</td>
<td>{{$result->gloss}}</td>
<td>{{$result->sequence}}</td>
<td>{{$result->lemma_id_fk}}</td>
<td>{{$result->derivatives_id}}</td>
<td>{{$result->etymology_id}}</td>
<td>{{$result->meaning}}</td>
<td>{{$result->semantic_category}}</td>
<td>{{$result->semantic_field}}</td>
<td>{{$result->simplicity_score}}</td>
<td>{{$result->age_score}}</td>
<td>{{$result->borrowed_score}}</td>
<td>{{$result->description}}</td>
<td>{{$result->typical_context}}</td>
<td>{{$result->representation}}</td>
<td>{{$result->sub_code}}</td>
</tr>
@endforeach
</tbody>
</table>==编辑#2 ==
只是在这里做了一些其他的测试,并意识到
$results = DB::table('lemma')
->join('etymology', 'lemma.id', '=', 'etymology.lemma_id_fk')
->join('gloss', 'etymology.id', '=', 'gloss.etymology_id')
->join('wold_meanings', 'gloss.gloss', '=', 'wold_meanings.meaning')
->select(DB::raw('lemma.*'))
->where('gloss.gloss', 'like', '%flower%')
->get();它只给出了表引理的结果,即忽略了联接。因此,我从<td>{{$result->id}}</td>收到视图表中的结果,直到<td>{{$result->pageid}}</td>,而不是其余的。
发布于 2017-04-11 11:03:38
从链中删除对select()的调用将使其工作。
select方法表示查询的SELECT部分。这样,编写以下内容:
->select(DB::raw('lemma.*'))这和:
SELECT lemma.* FROM ...换句话说,您只是从lemma表中选择列。
但是,由于您也希望从其他表中选择列,因此忽略链的这一部分将使查询生成器回退到其默认行为,这与以下相同:
SELECT * FROM ...如果您想要明确您所选择的内容,那么这样做将实现同样的目标:
->select('*')https://stackoverflow.com/questions/43323439
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