简单的php + mysql从数据库打印变量

Question

抱歉打扰大家，但我真的为此感到挣扎：

我很好地连接到数据库，然后尝试以下mysql语句：

$query1 = "select row1 from mydatabase where row2 = $Name ";
$answer1 = mysql_query($query1);

但是，稍后尝试时会出现以下几行：

echo $answer1;

我只得到null :(

有人可以给我任何建议吗？

编辑：SQL登录名：

mysql_connect("correct", "username", "password");

mysql_select_db("dbname") or die(mysql_error());

Answer 1

您所做的一切都是正确的，您只需要像这样获取数据：

$query1 = "select row1 from mydatabase where row2 = $Name ";
$answer1 = mysql_query($query1);

while($data= mysql_fetch_array($answer1)){
    echo $data['row1'];
}

这是一个完整的答案，我会根据您的需要进行调整;）

<?php
    //Connect to your database
    $con=mysqli_connect("db_hostname","db_user","db_password","db_name");

    // Check connection
    if (mysqli_connect_errno())
    {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

    //Value of the row to select
    $row2 = 'some value';

    //Make select query
    $result = mysqli_query($con, "SELECT row1 FROM MyTable WHERE row2='$row2'");

    //Fetch datas
    while($row = mysqli_fetch_array($result))
    {
        echo $row['row1'];
        echo "<br>";
    }

    //Close database
    mysqli_close($con);
?>

祝好运 ：）

Answer 2

尝试使用MySQLi_*代替MySQL_*函数，并将连接变量传递给函数调用。

如果这不起作用，那么您可能希望通过启用所有错误报告并转储全局作用域来尝试进一步调试。

<?php
error_reporting(E_ALL); // Show all errors & warnings

$conn = mysqli_connect("server", "username", "password");
mysqli_select_db($conn, "dbname") or die(mysql_error());

$sql1 = "SELECT `row1` FROM `mydatabase` WHERE `row2` = '".$Name."';";
$query1 = mysqli_query($conn, $sql1);
$answer1 = mysqli_fetch_assoc($query1);

var_dump($GLOBALS); // Dumps all variables in the global scope
?>

Answer 3

在$answer1= mysql_query($query1);之后添加它$answer1= mysql_query($query1);

while ($row = mysql_fetch_assoc($answer1)) {
   // echo data
   echo $row['row1'];
}