Php 在in子句Mysql中传递字符串变量
我有这段代码,我需要在in子句中传递一个变量,以便从数据库中捕获所有相应的记录。问题是,当有多条记录时,它只返回第一条记录。。问题出现在第二个查询中:Php 在in子句Mysql中传递字符串变量,php,mysql,Php,Mysql,我有这段代码,我需要在in子句中传递一个变量,以便从数据库中捕获所有相应的记录。问题是,当有多条记录时,它只返回第一条记录。。问题出现在第二个查询中: // FIRST QUERY $query0 = "SELECT articoli.id FROM articoli WHERE articoli.visibile = ? AND articoli.genere2 = ? ORDER BY artic
// FIRST QUERY
$query0 = "SELECT articoli.id
FROM articoli
WHERE articoli.visibile = ?
AND articoli.genere2 = ?
ORDER BY articoli.id ASC";
$stmt0 = mysqli_stmt_init($con);
mysqli_stmt_prepare($stmt0,$query0);
mysqli_stmt_bind_param($stmt0,'ii',$visibile,$id);
mysqli_stmt_bind_result($stmt0,$rows0['id']);
mysqli_stmt_execute($stmt0);
while (mysqli_stmt_fetch($stmt0)){
$array_id[] = $rows0['id'];
}
$array_id = implode(',', $array_id);
print_r($array_id);
echo gettype($array_id)."<br />";
$immagini = array();
// SECOND QUERY:
$query1 = "SELECT articoli.id AS id_articoli,
articoli.titolo,
articoli.descrizione,
galleria.id AS id_galleria,
group_concat(galleria.foto) as immagini
FROM articoli
LEFT JOIN galleria
ON articoli.id = galleria.rif_id
WHERE FIND_IN_SET(articoli.id,?) // => Initially it was 'WHERE articoli.id IN ?'
AND articoli.visibile = ?
GROUP BY articoli.id
ORDER BY articoli.id";
$stmt1 = mysqli_stmt_init($con);
mysqli_stmt_prepare($stmt1,$query1);
mysqli_stmt_bind_param($stmt1,'ii',$array_id,$visibile);
// INIZIALIZZO LA CONNESSION
$stmt1 = mysqli_stmt_init($con);
// PREPARARE QUERY
mysqli_stmt_prepare($stmt1,$query1);
// LEGO I PARAMETRI
mysqli_stmt_bind_param($stmt1,'ii',$array_id,$visibile);
mysqli_stmt_bind_result($stmt1,$rows1['id_articoli'],
$rows1['titolo'],
$rows1['descrizione'],
$rows1['id_galleria'],
$rows1['immagini']);
// ESEGUO LA QUERY
mysqli_stmt_execute($stmt1);
while (mysqli_stmt_fetch($stmt1)){
$html .= "<p>$rows1[id_articoli]</p>";
$html .= "<p>$rows1[titolo]</p>";
$html .= "<p>$rows1[descrizione]</p>";
$html .= "<p>$rows1[id_galleria]</p>";
$html .= "<p>$rows1[immagini]</p>";
//$immagini = explode(',', $rows1['immagini']);
//$html .= "<p>$immagini[0]</p>";
}
return $html;
mysqli_stmt_close($stmt1);
在上面的代码中,在第一个查询中,我有一个id列表,我将其放入一个数组中,然后我将数组内爆为一个字符串。然后,使用print\r,我可以查看所有记录。使用echo gettype,它告诉我变量是字符串。然后我在in子句中使用变量,但它总是只返回第一条记录。。然后我在谷歌上发现了FIND_IN_SET命令,我修改了代码并尝试了它,但没有改变。它总是只给我返回第一张唱片。。我怎样才能解决这个问题?我必须使用一个字符串,或者我可以向查询传递一个数组,最好使用IN子句或FIND_IN_SET子句?摆脱第一个查询,只在第二个查询中使用这些参数,而不需要FIND_IN_SET来匹配ID
$query1 = "SELECT articoli.id AS id_articoli,
articoli.titolo,
articoli.descrizione,
galleria.id AS id_galleria,
group_concat(galleria.foto) as immagini
FROM articoli
LEFT JOIN galleria
ON articoli.id = galleria.rif_id
WHERE articoli.visibile = ?
AND articoli.genere2 = ?
GROUP BY articoli.id
ORDER BY articoli.id";
$stmt1 = mysqli_stmt_init($con);
mysqli_stmt_prepare($stmt1,$query1);
mysqli_stmt_bind_param($stmt1,'ii',$visibile, $id);
为什么不把这两个查询合并起来呢?实际上,为什么你甚至需要两个查询呢?只要把articoli.visiblile=?和articoloi.genere2=?进入第二个问题,我没想过。。。这很简单。。。好的,我试试看,然后我告诉你:-谢谢!这很简单,我确实需要休息一下。。。谢谢如果你把它当作解决方案张贴,我会投你的票