无法在php中使用mysql查询插入数据

我正在尝试使用表单将一些数据插入数据库。填写表格后，我使用method=“POST”获取数据。我没有语法错误，尽管我无法将表单中的数据插入数据库

以下是php的一些代码：

<?php

    // Connects to your Database
    $con = mysql_connect("localhost", "root", "134711Kk", "eam3");

    error_reporting(E_ALL);
    ini_set('display_errors', 1);

    //This is the directory where images will be saved
    $target = "dokupload/";
    $target = $target . basename( $_FILES['photo']['name']);
    echo $target;
    echo "<br>";

    $name=$_POST['nameMember'];
    $bandMember=$_POST['bandMember'];


    if(!$con)
    {
        die("Connection Failed".mysql_error());     
    }

    echo "$name" . " " . "$bandMember" . "<br/>";

    $sql = "INSERT INTO Grammateia(id, idruma, tmhma) VALUES ('1', '$name', '$bandMember');";   

    $some = mysql_query($sql, $con);

    $request = mysql_query("SELECT * FROM eam3.Grammateia;", $con);

    while ($row = mysql_fetch_array($request))
    {   
        extract($row);
        echo "$id" . " " . "$idruma" . " " . "$tmhma" . "<br/>";
    }


    .
    .
    .

    mysql_close($con);
?> 

---

尝试删除“；”从你的疑问

这应该是您最好的资源
使用mysqli而不是mysql，因为它已被弃用

编写
mysqli\u查询（$con，$sql）
not
mysqli\u查询（$sql，$con）

试试下面的代码

<?php

    // Connects to your Database
    $con = mysqli_connect("localhost", "root", "134711Kk", "eam3");

    error_reporting(E_ALL);
    ini_set('display_errors', 1);

    //This is the directory where images will be saved
    $target = "dokupload/";
    $target = $target . basename( $_FILES['photo']['name']);
    echo $target;
    echo "<br>";

    $name=$_POST['nameMember'];
    $bandMember=$_POST['bandMember'];


    if(!$con)
    {
        die("Connection Failed".mysqli_error());     
    }

    echo "$name" . " " . "$bandMember" . "<br/>";

    $sql = "INSERT INTO Grammateia(id, idruma, tmhma) VALUES ('1', '$name', '$bandMember');";   

    $some = mysqli_query($con,$sql);

    $request = mysqli_query($con,"SELECT * FROM eam3.Grammateia");

    while ($row = mysqli_fetch_array($request))
    {   
        extract($row);
        echo "$id" . " " . "$idruma" . " " . "$tmhma" . "<br/>";
    }
    mysqli_close($con);
?> 

您的示例不是插入数据的推荐方法（为了安全起见，您应该真正使用
mysqli->prepare
）。但是，要修复这里的问题：

$sql = "INSERT INTO Grammateia(id, idruma, tmhma) VALUES ('1', '" . $name . "', '" . $bandMember . "');";   

现在更改您的密码。看看PHP的mysql_uAPI、sql注入和prepared语句的弃用。为什么要指向弃用的API？不过，公平地说，我想这个链接至少说它已经被弃用了。
$sql = "INSERT INTO Grammateia(id, idruma, tmhma) VALUES ('1', '" . $name . "', '" . $bandMember . "');";