PHP-如何第一次插入表单数据和第二次更新相同的数据

问题描述：

I have a form having input field name (id,settings,value) with submit button. Now, I am inserting form data into database now I want that when from data inserting first time then how to update same form data second time Wthout insert data..

    <?php 
    if(isset($_POST['save']))
    {
    include "connection.php" ;

    $id = $_POST['id'] ;
    $name = $_POST['name'] ;
    $value   = $_POST['value'] ;

$query="create table if not exists settings(id int(10),name varchar(50),value varchar(100))";
$results=mysql_query($query) or die("QUERY FAILED 1:".mysql_error());   

$query="INSERT INTO settings VALUES('$id','$name','$value')"; 
$results=mysql_query($query) or die("QUERY FAILED 2:".mysql_error());


 $query="update settings set value='$value' where id='$id'";
 $results=mysql_query($query) or die("QUERY FAILED 3:".mysql_error());



     echo $settings ;

    }

我有一个表单，其中包含输入字段名称（id，settings，value）和提交按钮。现在，我正在将表格数据插入到数据库中我现在想要从第一次插入数据时那么第二次如何更新相同的表格数据Wthout插入数据.. p>

 ＆lt;？  php 
 if（isset（$ _ POST ['save']））
 {
 include“connection.php”; 
 
 $ id = $ _POST ['id']; 
 $ name = $ _POST  ['name']; 
 $ value = $ _POST ['value']; 
 
 $ query =“如果不存在则创建表设置（id int（10），name varchar（50），value varchar（100  ））“; 
 $ results = mysql_query（$ query）或die（”QUERY FAILED 1：“。mysql_error（））;  
 
 $ query =“INSERT INTO设置VALUES（'$ id'，'$ name'，'$ value'）”;  
 $ results = mysql_query（$ query）或die（“QUERY FAILED 2：”。mysql_error（））; 
 
 
 $ query =“update settings set value ='$ value'其中id ='$ id  '“; 
 $ results = mysql_query（$ query）或die（”QUERY FAILED 3：“。mysql_error（））; 
 
 
 
 echo $ settings; 
 
} 
  code  >  pre> 
  div>

答

You need query the table to check whether the value is exists or not. If the record is not exists, then insert it, else update it.

<?php
    if(isset($_POST['save']))
    {
    include "connection.php" ;
    $id = $_POST['id'] ;
    $name = $_POST['name'] ;
    $value   = $_POST['value'] ;

    $query="create table if not exists settings(id int(10),name varchar(50),value varchar(100))";
    $results=mysql_query($query) or die("QUERY FAILED 1:".mysql_error());


    // pass the name and check whether the value is exists or not,
    $result = mysql_query("SELECT * FROM settings WHERE name = '".$name."'");
    $number_of_rows = mysql_num_rows($result);

    // if number of rows is 0, then insert it. else update it
    if($number_of_rows == 0){
        $query="INSERT INTO settings VALUES('$id','$name','$value')";
    } else{
        // get the row id
        $data_row = mysql_fetch_row($result);
        $record_id = $data_row[0];
        $query="UPDATE settings SET value='".$value."' WHERE id = '".$record_id."'";
    }

    $results=mysql_query($query) or die("QUERY FAILED 2:".mysql_error());

     echo "Record inserted/updated";
    }

PHP-如何第一次插入表单数据和第二次更新相同的数据

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