您想为以下查询结果创建一个关联数组:
$sql = "select s.*,di.dealsimage,
ctm.city,
l.location,
GROUP_CONCAT(DISTINCT cm.cuisine ORDER BY scr.cuisine_sequence_for_store) AS cui,
GROUP_CONCAT(DISTINCT rtm.restaurant_type ORDER BY srr.rest_type_sequence_for_store) AS restauranttype
from stores s
left join city_master ctm on s.city_id = ctm.city_id
left join locations l on s.location_id = l.location_id
left join store_cuisine_relation scr on s.store_id = scr.store_id
left join cuisine_master cm on scr.cuisine_id = cm.cuisine_id
left join store_resttype_relation srr on s.store_id = srr.store_id
left join restaurant_type_master rtm on rtm.rest_type_id = srr.rest_type_id
left join store_dealcat_relation sdr on s.store_id = sdr.store_id
left join deals_category_master dcm on dcm.deal_cat_id = sdr.deal_cat_id
left join deals_image di on di.`store_id` = s.store_id
where $condition1 s.is_active = 1 $condition2 group by (s.store_id) order by s.store_id";
//echo $sql;exit;
//echo $sql;exit;
$sqlex1 = mysqli_query($db,$sql);
$custom_count = @mysqli_num_rows($sqlex1); // it prints 28
while($result1 = mysqli_fetch_assoc($sqlex1)){
$dataArr = array_push_assoc($dataArr, 'store_id', $result1['store_id']);
$dataArr = array_push_assoc($dataArr, 'store_name', $result1['store_name']);
$dataArr = array_push_assoc($dataArr, 'store_logo', $result1['store_image_url']);
$dataArr = array_push_assoc($dataArr, 'deals_image', $result1['dealsimage']);
}
//echo count($dataArr);exit;
//echo $kl;exit;
//$result = array_merge_recursive($gpsArr,$dataArr);
function array_push_assoc($array, $key, $value){
$array[$key][] = $value;
return $array;
}
查询返回28结果但是当我尝试回显$dataArr
的计数时,它总是打印4。我做错了什么?我怎样才能做到这一点?
提前致谢
答案 0 :(得分:0)
通过您的功能array_push_assoc()
创建包含4个键store_id
,store_name,
store_logo , and
deals_image`的数组,每个键有28行
试试这个:
dataArr = array();
while($result1 = mysqli_fetch_assoc($sqlex1))
{
$dataArr[] = array('store_id'=>$result1['store_id'],
'store_name'=>$result1['store_name'],
'store_logo'=>$result1['store_image_url'],
'deals_image'=>$result1['dealsimage']);
}
这将创建具有28个关联数组的数组
答案 1 :(得分:0)
您的代码应更改为:
while($result1 = mysqli_fetch_assoc($sqlex1))
{
array_push($dataArr, array ('store_id'=> $result1['store_id'], 'store_name'=> $result1['store_name'], 'store_logo'=> $result1['store_image_url'], 'deals_image'=> $result1['dealsimage']));
}