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[英]PHP MySQL Update not working when using variable in WHERE clause
[英]PHP MySQL result issue when using a variable in the WHERE Clause
我的PHP MySQL查詢存在一個很奇怪的問題,我試圖返回具有正確rname,rcity和rstate值的行。
$sql = "SELECT * FROM `images` WHERE rname = '" . $rname . "' AND rcity = '" . $rcity . "' AND rstate = '" . $rstate . "'";
當我運行該查詢時,它僅返回0個結果。 但是,經過一番嘗試之后,如果我僅在WHERE子句中使用rname和rstate,它將返回結果。
$sql = "SELECT * FROM `images` WHERE rname = '" . $rname . "' AND rstate = '" . $rstate . "';
完美的作品。 因此,當我嘗試僅在WHERE子句中使用rcity時。
$sql = "SELECT * FROM `images` WHERE rcity = '" . $rcity . "'";
返回0個結果。 因此查詢的rcity部分出了問題。 如果我將值而不是$ rcity變量硬寫到查詢中,則會拉出結果。 我仔細檢查了一下,以確保聲明了$ rcity,它具有正確的值,等等。
我還在數據庫中創建了另一個測試表,以檢查它在數據庫方面是否存在問題。 哪個問題仍然存在。
這是getQuery()函數的完整代碼
private function getQuery($data){
// Takes raw data and creats image(s) query to search for listing resort...
$listing = $data['listing'];
$rname = $data['rname'];
$rcity = $data['rcity'];
$rstate = $data['rstate'];
$query = "SELECT * FROM `test` WHERE rname = '" . $rname . "' AND rcity = '" . $rcity . "' AND rstate = '" . $rstate ."'";
return $query;
}
這是我的數據庫類
class db {
public function __construct(){
$this->server = DB_SERVER;
$this->user = DB_USER;
$this->Pass = DB_PASS;
$this->Database = DB_Database;
}
protected function connect(){
return mysqli_connect($this->server, $this->user, $this->Pass, $this->Database);
}
public function query($sql){
$conn = $this->connect();
$query = $conn->query($sql);
if($query == false) {
throw new Exception("Query failed:".PHP_EOL.$conn->error.PHP_EOL.$sql);
}
if($query->num_rows == 0) {
// need E_NOTICE errors enabled to see this,
// on screen if display_errors is on, else in PHP error log
trigger_error("Query returned 0 rows:".PHP_EOL.$sql);
}
$result = array();
while ($row = $query->fetch_assoc()){
$result[] = $row;
}
return $result;
}
}
我像這樣在類__construct函數中調用查詢
$con = new db;
$sql = $this->getQuery($data);
$result = $con->query($sql);
我認為問題可能出在語法或mysql篩選上。 嘗試將PDO與bindParam方法一起使用
$sql = "SELECT * FROM `images` WHERE rname = '$rname' AND rcity = '$rcity' AND rstate = '$rstate'";
嘗試實現這一點,即直接在'
(撇號)之間使用變量,它應該可以完美地工作
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