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我試圖從另一個數據庫值中使用ID獲取一些數據。這裏是我的代碼:PHP MySQLi準備語句獲取行,但不顯示數據
$query = "SELECT id,uid,product,jpgID,saledatetime,quantity
FROM sales
WHERE printed = ?
AND DATE_FORMAT(saledatetime, '%Y/%m/%d') = ?";
if ($stmt = $mysqli->prepare($query)) {
$printed = 0;
$stmt->bind_param("is",$printed,$session_date);
$stmt->execute();
$stmt->store_result();
//Get result
$stmt->bind_result($result_id,$result_uid,$result_product,$result_jpgID,$result_saledatetime,$result_quantity);
//Number of rows returned
$count_rows = $stmt->num_rows;
if ($stmt_fetch = $mysqli->prepare("SELECT jpg,udate,imageset FROM images WHERE id = ?")) {
//Fetch image information
while ($stmt->fetch()) {
$stmt_fetch->bind_param('i',$result_jpgID);
$stmt_fetch->execute();
$stmt_fetch->store_result();
$stmt_fetch->bind_result($jpg,$udate,$imageset);
echo $stmt_fetch->num_rows."<br />";
//Create array's with information
$print[$result_product][$result_jpgID]["quantity"] = $result_quantity;
$print[$result_product][$result_jpgID]["location"] = $jpg;
}
}
$stmt->close();
print_r($print);
}
的echo $stmt_fetch->num_rows."<br />";
顯示1每次循環的時間,所以就找到了匹配的行。但對於沒有綁定的變量的一些原因,沒有任何價值......這是顯示結果:
1
1
1
Array
(
[KR] => Array
(
[137] => Array
(
[quantity] => 1
[location] =>
)
[138] => Array
(
[quantity] => 1
[location] =>
)
)
[LR] => Array
(
[138] => Array
(
[quantity] => 1
[location] =>
)
)
)
並且您完全確定查看jpg,udate,imageset的第二個查詢實際上與手動運行時提供的ID一起使用?即。在phpmyadmin或控制檯?另外,對原始查詢(我甚至不知道)進行連接並不容易,然後只是循環一個結果集,而不是將這2個結合起來使用 – Tularis
@Tularis,嗯它不運行如果我手動定義標識..:S這是我知道這樣做的唯一途徑,我將如何結合這些查詢呢? –
還添加了第一個查詢。 –