0

我一直在努力做到这一点，但作为其他问题，我就是想不通。我一直在阅读可能的解决方案，但没有一个符合我的代码，或者如果他们这样做，我不知道如何或在哪里使用它们。

我有一个用户发送记录的数据库。该数据库由几个表组成，其中包含以下“姓名、姓氏、电话”。如果这些值中的任何一个是重复的，我希望我的代码能够识别并忽略表单的提交，如果所有这些值已经存在于 DB 上。

这是我的代码：

        <?php

        $con = mysql_connect("HOST","USER","PASS");
        if (!$con)
          {
          die('Could not connect: ' . mysql_error());
          }

        mysql_select_db("testdb", $con);
        $sql="INSERT INTO people (Name, LastName, Phone)
        VALUES
        ('$_POST[Name]','$_POST[LastName]','$_POST[Phone]')";

        if (!mysql_query($sql,$con))
          {
          die('Error: ' . mysql_error());
          }
        
        echo "Record Added";

        mysql_close($con);
        ?>
4

3 回答 3

2

该mysql_*功能现在已全部弃用，不应使用。更改您的代码以执行以下操作：

//Set up a PDO connection to MySQL
$host = 'host_name';
$dbname = 'database_name';
$user = 'user_name';
$pass = 'user_pass';
try
{
    $DB = new PDO("mysql:host=$host;dbname=$dbname", $user, $pass); 
}
catch(PDOException $e)
{  
    echo $e->getMessage();  
}

//Determine whether the appropriate values have been passed.
if(isset($_POST['Name']))
{
    $name = $_POST['Name'];
}
else
{
    echo "You must provide a name!";
        exit; //This may not be what you want to do here, it's an example action
}

if(isset($_POST['LastName']))
{
        $name = $_POST['LastName'];
}
else
{
    echo "You must provide a last name!";
        exit; //This may not be what you want to do here, it's an example action
}

if(isset($_POST['Phone']))
{
        $name = $_POST['Phone'];
}
else
{
    echo "You must provide a phone number!";
        exit; //This may not be what you want to do here, it's an example action
}


//Set up the query using anonymous values
$sql="INSERT INTO people (Name, LastName, Phone) VALUES ('?','?','?')";
$sth = $DB->prepare($sql);

try
{
        //Attempt to execute the insert statement
        $sth->execute(array($_POST[Name], $_POST[LastName], $_POST[Phone]));
        echo "Record Added";

}
catch(PDOException $e)
{
        //If the insert failed, then you can handle the error, and determine
        //what further steps need to be taken.
        echo "Record Not Added";
}

这是另一个具有类似设置的问题，可能对您也有用：

https://stackoverflow.com/a/10414922/1507210

于 2012-10-31T17:26:47.567 回答
0

插入前在表中搜索

 <?php

        $con = mysql_connect("HOST","USER","PASS");
        if (!$con)
          {
          die('Could not connect: ' . mysql_error());
          }

        mysql_select_db("testdb", $con);

        $name =  mysql_real_escape_string($_POST[Name]);
        $LastName=  mysql_real_escape_string($_POST[LastName]);
        $Phone=  mysql_real_escape_string($_POST[Phone]);

         $search_res=mysql_query("SELECT * from people where Name='$Name' OR LastName='$LastName' OR Phone='$Phone'");

        if(mysql_num_rows($search_res) < 1){

             $sql="INSERT INTO people (Name, LastName, Phone)
             VALUES
                ('$Name','$LastName','$Phone')";

               if (!mysql_query($sql,$con))
              {
              die('Error: ' . mysql_error());
              }

        echo "Record Added";
       }else{
          echo "User Already exits";
       }

        mysql_close($con);
        ?>
于 2012-10-31T17:32:12.253 回答
0

试试这个简单的解决方案

$result = mysql_query("SELECT * FROM TABLE WHERE Column = 'value' ");

if( mysql_num_rows($result) < 1) {
    mysql_query("INSERT INTO table (column) VALUES ('value') ");
}
于 2012-10-31T17:45:12.040 回答