我正在使用 postgres 擴展在 php 中準備一個語句。然后我嘗試使用pg_fetch_object.從準備好的語句中沒有返回任何行，但它應該。我還收到以下警告：警告：pg_prepare(): Query failed: ERROR: Prepared statement "parking" already exists in C:\xampp\htdocs\map\MapMarkers.php on line 33$devices_query = pg_query($conn, "SELECT applications.name category, devices.* FROM app_as.application applications, app_as.device devices WHERE applications.id = devices.application_id");//variables are bound in the loop$parking_pst = pg_prepare($conn, "parking", "SELECT distinct on (name) name,application_name,longitude,latitude,parking_car_status status,received_at FROM V_DEVICE_PARKING WHERE name = $1");while ($devices = pg_fetch_object($devices_query)) { pg_execute($conn, "parking",[$devices->name]); while ($parking = pg_fetch_object($parking_pst)) { $devices->parking_car_status = $parking->parking_car_status; } $data['DEVICES'][] = $devices;}
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TA貢獻1824條經驗 獲得超5個贊
您收到此錯誤/警告是因為您正在pg_fetch_object()通話pg_prepare()。本質上，每次在while()循環中迭代時，您都會再次調用pg_prepare()，無意中嘗試創建一個新的準備好的語句，稱為“停車”。
您真正要做的是獲得結果，pg_execute()然后使用以下方法提取結果pg_fetch_object()：
$devices_query = pg_query($conn, "SELECT applications.name category, devices.* FROM app_as.application applications, app_as.device devices WHERE applications.id = devices.application_id");
$prepare_result = pg_prepare($conn, "parking", "SELECT distinct on (name) name,application_name,longitude,latitude,parking_car_status status,received_at FROM V_DEVICE_PARKING WHERE name = $1");
// Maybe process $prepare_result as needed here
while ($devices = pg_fetch_object($devices_query)) {
$execute_result = pg_execute($conn, "parking", [$devices->name]);
while ($parking = pg_fetch_object($execute_result)) {
$devices->parking_car_status = $parking->parking_car_status;
}
$data['DEVICES'][] = $devices;
}
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