Php 通过连接库存和特征表,使用数组字符串显示库存的特征
我使用query1检索了stock\u id,并将其作为$row['stock\u id']使用。我想先将此功能保存在数组中,以显示此库存id的所有功能 这是我不工作的代码Php 通过连接库存和特征表,使用数组字符串显示库存的特征,php,mysql,join,implode,Php,Mysql,Join,Implode,我使用query1检索了stock\u id,并将其作为$row['stock\u id']使用。我想先将此功能保存在数组中,以显示此库存id的所有功能 这是我不工作的代码 'stock' table >> stock_id integer NOT NULL AUTO_INCREMENT, modal varchar(40) NOT NULL DEFAULT "", company varchar(40
'stock' table >>
stock_id integer NOT NULL AUTO_INCREMENT,
modal varchar(40) NOT NULL DEFAULT "",
company varchar(40) NOT NULL DEFAULT "",
bike_number varchar(10) NOT NULL DEFAULT "",
lair_id integer NOT NULL DEFAULT 0,
PRIMARY KEY(stock_id)
'feature' table >>
feature_id integer NOT NULL AUTO_INCREMENT,
feature varchar(40) NOT NULL DEFAULT "",
primary key (feature_id)
'stock_feature' linking table >>
stock_id integer NOT NULL,
feature_id integer NOT NULL,
primary key(stock_id, feature_id)
$query2='f.feature\u id=sf.feature\u id,其中sf.stock\u id='。$row['stock\u id'];
$result2=mysqli_query($db,$query2)或die(mysqli_error($db));
如果(mysqli_num_行($result2)>0){
$features=array();
而($row2=mysqli\u fetch\u assoc($result2)){
$features[]=$row2['feature'];
}
回显“”。内爆(',',$features)。“”;
}否则{
回声“无明显特征”;
}
$query2='SELECT feature FROM feature f JOIN stock_feature sf ON f.feature_id=sf.feature_id WHERE sf.stock_id='.$row['stock_id'];
$result2=mysqli_query($db, $query2) or die(mysqli_error($db));
if(mysqli_num_rows($result2)>0){
$features=array();
while($row2=mysqli_fetch_assoc($result2)){
$features[]=$row2['feature'];
}
echo '<td>' . implode (', ', $features) . '</td>';
}else{
echo '<td> No significant feature </td>';
}