Question

我正在创建一个由数据库组成的Web服务器，当输入代码时，它将检索数据库中的一行。

这是我的代码为什么输出这个：

“。$ conn-＆gt;错误;}}？php＆gt;

我的代码： -

<php
echo "HI";
$Hog_ID=filter_input(INPUT_POST, 'Hog_ID');

$host = "localhost";
$dbusername = "teambaboy";
$dbpassword = "pigislove";
$dbname = "teambaboy";

$conn = new mysqli ($host, $dbusername, $dbpassword, $dbname);

if (mysqli_connect_error()){
    die('Connect Error ('.mysqli_connect_errno().') '
    . mysqli_connect_error());
}

else{

   $sql = "SELECT (Animal_Number, Farm_code, Dam, Sire, Breed, Birth_date, Birth_weight, Date_worn, Wearing_weight, Feeds_given, Amount) FROM pig_registration WHERE Animal_Number = $Hog_ID";

    if (mysqli_query($conn, $sql)){
        echo $boom["Animal_Number"]. $boom["Farm_code"]. $boom["Dam"]. $boom["Sire"]. $boom["Breed"]. $boom["Breed"]. $boom["Birth_date"]. $boom["Birth_weight"]. $boom["Date_worn"]. $boom["Wearing_weight"]. $boom["Feeds_given"]. $boom["Amount"].;
    }
    else{
        echo "Error: ". $sqL ."<br>". $conn->error;
    }

}


?php>

Answer 1

您错过了while循环以及$boom变量未声明

 $sql = "SELECT (Animal_Number, Farm_code, Dam, Sire, Breed, Birth_date, Birth_weight, Date_worn, Wearing_weight, Feeds_given, Amount) FROM pig_registration WHERE Animal_Number = $Hog_ID";

$result =mysqli_query($conn, $sql);
while($boom = mysqli_fetch_assoc($result)){ // mysqli_fetch_assoc() Returns an associative array of strings that corresponds to the fetched row, or FALSE if there are no more rows.

    echo $boom["Animal_Number"]. $boom["Farm_code"]. $boom["Dam"]. $boom["Sire"]. $boom["Breed"]. $boom["Breed"]. $boom["Birth_date"]. $boom["Birth_weight"]. $boom["Date_worn"]. $boom["Wearing_weight"]. $boom["Feeds_given"]. $boom["Amount"];
}
}

另外，我建议你使用prepared queries or PDO来帮助你防止sql注入

Answer 2

你应该试试这个 -

<?php
$Hog_ID=filter_input(INPUT_POST, 'Hog_ID');
//$Hog_ID=2; for test pass static Hog_ID

$host = "localhost";
$dbusername = "teambaboy";
$dbpassword = "pigislove";
$dbname = "teambaboy";

$conn = new mysqli ($host, $dbusername, $dbpassword, $dbname);

if (mysqli_connect_error()){
    die('Connect Error ('.mysqli_connect_errno().') '
    . mysqli_connect_error());
}else{

    $sql = "SELECT Animal_Number, Farm_code, Dam, Sire, Breed, Birth_date, Birth_weight, Date_worn, Wearing_weight, Feeds_given, Amount FROM pig_registration WHERE Animal_Number = $Hog_ID";
    $result = mysqli_query($conn, $sql);
    $boom = mysqli_fetch_assoc($result);

    if(!empty($boom)){
    //print_r($boom);

        echo $boom['Animal_Number'].$boom['Farm_code'].$boom['Dam'].$boom['Sire'].$boom['Breed'].$boom['Birth_date'].$boom['Birth_weight'].$boom['Date_worn'].$boom['Wearing_weight'].$boom['Feeds_given'].$boom['Amount'];
    }else{
        echo "Error: ". $sqL ."<br>". $conn->error;
    }

}
?>

PHP Mysql打印出一行

2 个答案: