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php - 数据库错误 Cakephp

转载 作者:行者123 更新时间:2023-11-30 22:42:24 27 4
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保存数据时出现数据库错误

错误

SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ') GROUP BY s.state, s.indicator_datum_id) AS LatestStateUpdate ON' at line 4

SQL 查询:

SELECT IndicatorDatum.id, IndicatorDatumUpdate.*, (CONCAT(IndicatorDatum.indicator_id,'_',IndicatorDatum.report_year_id)) 
AS IndicatorDatum__indexKey FROM astest.indicator_data AS IndicatorDatum
INNER JOIN (SELECT u.indicator_datum_id, MAX(u.created) as update_date, s.state

FROM indicator_datum_states s
INNER JOIN indicator_datum_updates u on u.id = s.indicator_datum_update_id
WHERE s.indicator_datum_id IN () GROUP BY s.state, s.indicator_datum_id) AS LatestStateUpdate ON (IndicatorDatum.id = LatestStateUpdate.indicator_datum_id AND IndicatorDatum.state = LatestStateUpdate.state)
INNER JOIN astest.indicator_datum_updates AS IndicatorDatumUpdate ON (IndicatorDatumUpdate.indicator_datum_id = LatestStateUpdate.indicator_datum_id AND IndicatorDatumUpdate.created = LatestStateUpdate.update_date)
WHERE IndicatorDatum.id = (NULL)

我想知道我的代码中有什么错误(阅读食谱后)我理解传递给保存方法的数组应该是这样的

array(
(int) 0 => array(
'value' => '',
'id' => '',
'indicator_id' => '283',
'report_year_id' => '7',
'Assurance' => array(
(int) 0 => '1',
(int) 1 => '2',
(int) 2 => '3',
(int) 3 => '4',
(int) 4 => '5'
)
),
(int) 1 => array(
'value' => '',
'id' => '',
'indicator_id' => '283',
'report_year_id' => '6',
'Assurance' => ''))

但是当我调试我的代码时,我发现数据传递给了保存方法:

   array(
'currentOrg' => array(
'id' => '40'
),
'IndicatorDatum' => array(
'$cn' => array(
'id' => '',
'comment' => '',
'reference' => ''
),
(int) 0 => array(
'value' => '',
'Assurance' => ''
),
(int) 1 => array(
'value' => '',
'Assurance' => ''
),
(int) 2 => array(
'value' => '',
'Assurance' => ''
),
(int) 3 => array(
'value' => '',
'Assurance' => ''
),

我的表单:

echo $this->DashboardForm->create('IndicatorDatum',array(
'url' => array('controller'=>'indicator_data','action'=>'edit_group', 'dashboard'=>true, $thisGroup['IndicatorGroup']['id']),
'novalidate'=>true
));
<? $cn = 0;
foreach ($years as $year) :
echo $this->Form->hidden("IndicatorDatum.$cn.id");
echo $this->Form->hidden("IndicatorDatum.$cn.state");
echo $this->Form->input("IndicatorDatum.$cn.indicator_id",array(
'type'=>'hidden', 'default'=>$iid
));
echo $this->Form->input("IndicatorDatum.$cnt.report_year_id",array(
'type'=>'hidden', 'default'=>$yid
));
echo $this->Form->input('IndicatorDatum.$cn.value');
echo $this->Form->input("IndicatorDatum.$cn.Assurance", array('style'=>'width: 165px;',
'type'=>'select',
'multiple'=>true, 'options' => $assurances, 'selected' => $selected,'label' => false));
$cn++;
endforeach;

echo $this->Form->submit(__('Save All'));

?>

我的 Controller :

if ($this->request->is('post')|| $this->request->is('put')) {
$data = debug($this->request->data);
if ($this->IndicatorDatum->saveAll($this->request->data)) {
$this->Session->setFlash(__('The indicator data has been saved'), 'flash/success');
$this->redirect(array('action'=>'edit_group',$group_id));
} else {
$this->Session->setFlash(__('The indicator data could not be saved. Please verify the fields highlighted in red and try again.'), 'flash/error');
}
}

最佳答案

您不应该通过任何未在 select 中指定的列,在这种情况下 s.state 和 s.indicator_datum_id 不会出现在 select 子句中。

关于php - 数据库错误 Cakephp,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/30755359/

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