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我需要使用從數據庫中獲取的某些數據填充數組。我的數據庫表如下所示:使用數據庫中的列填充數組
Tablename
ID | PROFILEID | PAGEID | VOTE
------------------------------------
1 | 1563187610 | /example.php| 1
2 | 1563187610 | /example.php| 2
3 | 1946357685 | /example.php| 1
------------------------------------
隨着時間一天天的代碼,我試圖用我總是得到一個數組,看起來像:Array ()
這是我用填充數組代碼:
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "test";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT FROM `Tablename`";
$result = mysql_query($sql);
$var = array();
while ($row = mysql_fetch_array($result)) {
$var[] = $row['PROFILEID'];
}
print_r($var);
$conn->close();
?>
UPDATE 1
$sql = "SELECT * FROM `Tablename`";
$result = mysqli_query($sql, $conn);
$var = array();
while ($row = mysqli_fetch_array($result)) {
$var[] = $row['PROFILEID'];
}
print_r($var);
仍然會導致Array()問題。
如果我把「or die(mysqli_error($conn))
」它沒有說什麼,我有一個空白屏幕
解決
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "test";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT * FROM `Tablename`";
$result = $conn->query($sql);
$var = array();
if ($result->num_rows > 0) {
// output data of each row
while ($row = mysqli_fetch_array($result)) {
$var[] = $row["PROFILEID"];
}
} else {
echo "0 results";
}
print_r ($var);
$conn->close();
您需要在您的查詢的佔位符SELECT * FROM'Tablename' –
你混合'mysql_ *'和' mysqli_ *'函數。這是行不通的。 –
不要以爲您的查詢會起作用。您還應該添加錯誤檢查,例如'或die(mysqli_error($ conn))'到您的查詢中。或者您可以在當前的錯誤日誌中找到問題。 –