PHP从mysql NOT DISPLAYING中选择一个数据

Question

我正在尝试显示或返回所选用户的日期值

function mysql_select_one_data($database, $table, $data, $country, $phone){  
$conn = mysql_connect("localhost", "root", "test123");

    mysql_select_db($database, $conn) or die (mysql_error());
    $query = "SELECT DATE FROM Temp_Users WHERE Country = '+94' AND Phone = '1234539543';";

    $result = mysql_query($query) or die (mysql_error());

    if (mysql_num_rows($result) > 0) {
        // output data of each row
        while($row = mysql_fetch_assoc($result)) {
            echo "Date: " . $row["Date"]. "<br>";
        }
    } else {
        echo "0 results";
    }

    return $result;
    }

结果：

Date:
Resource id #5

但是当我在phpmyadmin中执行查询时，它会返回

我是Php的新手，我不知道自己做错了什么，谢谢你的帮助。

Answer 1

更改此

$query = "SELECT DATE FROM Temp_Users WHERE Country = \"+94\" AND Phone = \"1234539543\";";

到这个

$query = "SELECT DATE FROM Temp_Users WHERE Country = '+94' AND Phone = '1234539543';";

将echo命令更改为echo "Date: $row['DATE']<br>"