简体   繁体   English

我正在尝试从数据库中提取数据,该数据将从数据库中的两个不同表中提取,错误在$ returns = mysql_query中

[英]I'm trying to pull data from database where the data will be pulled from two different tables in the db,error is in $returns = mysql_query

I'm trying to pull data from two different tables in the database and I keep on having error. 我正在尝试从数据库中的两个不同表中提取数据,并且一直出错。 when I run the code it displays nothing 当我运行代码时,它什么也不显示

Here is my code: 这是我的代码:

<?php 
$st="";
$sc="";

    if (isset($_POST['submit']) && $_POST['submit'] == "submit"){
       $st=$_POST['s-t'];
       $sc= "WHERE nums like \"%$st%\"  or d_n like \"%st%\" ";
    }


    if (isset ($_GET['look']) && $_GET ['look'] != "") {
       $st=$_GET['look'];
       $sc = "WHERE nums like \"%$st%\"  or d_n like \"%st%\" ";
     }

    $db = mysql_db_connect();
    $rpp = 50;

    if (isset($_GET['page'])){

    $pn = $_GET['page'];
    $begin = ($rpp*$pn)-$rpp;
    $finish=($rpp*$pn);
    $finish=50;
} 
else {
    $pn =1;
    $begin = 0;
    $finish =$rpp;
} 

    $returns = mysql_query ("SELECT * FROM  extenz $sc", $db);
    $numer_rows = mysql_num_rows($returns);
    $number_pages = ((int) ($numer_rows/$rpp)) +1;

    if (($number_rows % rpp) ==0){
    $number_pages=$number_pages -1;
}
$returns = mysql_query("SELECT nums, name , p_m FROM extenz, P_exten  left join p_exten on extenz $sc order by nums & P_m LIMIT $begin, $finish", $db);

?> 

i answered my question 我回答了我的问题

the error was in the following line of code 错误在以下代码行中

$returns = mysql_query("SELECT nums, name , p_m FROM extenz, P_exten  left join p_exten on extenz $sc order by nums & P_m LIMIT $begin, $finish", $db);

and the correction was 更正是

$returns = mysql_query("SELECT extenz.nums, extenz.name,p_exten.P_M FROM extenz LEFT JOIN ON  p_exten.extenz=extenz.nums $sc ORDER BY nums LIMIT $begin, $finish",$db);

暂无
暂无

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

相关问题 我正在尝试将数据从数据库中提取到选择框中 - I'm trying to pull data from database into selectboxes 我正在尝试使用从 php 数据库中提取的一些数据作为可点击 SVG 图片(javascript)中的信息 - I'm trying to use some data i pulled from a php database as information in a clickable SVG picture (javascript) 尝试使用where子句从两个不同的表中提取时出现MySQL查询错误 - Mysql query error when trying to extract from two different tables with where clause 从MySQL中的不同表中提取数据 - Pull out data from different tables in mysql mysql_fetch_array和mysql_query在来自两个不同表的调用中不返回任何内容 - mysql_fetch_array and mysql_query return nothing on a call from two different tables 我正在尝试从数据库中提取结果,并在从数据库中提取结果的链选择框中实用地选择结果 - I'm trying to pull results from a database and pragmatically select the results in a chain-select box that has it's results pulled from a database 在PHP / MYSQL中从同一数据库的两个不同表中选择数据 - Selecting data from two different tables of the same database in PHP/ MYSQL 使用SELECT查询从数据库中的两个不同表中获取数据 - Using a SELECT query to get data from two different tables in database 我一直在尝试使用 eloquent 从我的数据库中的两个不同表中检索数据 - i have been trying to retrive data from two different tables in my database using eloquent 从两个表中提取数据 - pull data from two tables
 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM