如何执行此查询？ [英] How to execute this query?

问题描述

我正在建立一个婚姻网站，现在我必须为每个用户发送一个示例个人资料，我想要一个独特的个人资料发送给每个用户，但无法找到如何做到这一点。

这是我的代码我做了什么。

 include（  db.php）; 
 $ query =   select * from matri_user_info，其中payment ='是'; 
 $ result = mysql_query（$ query，$ con）或 die（  Mysql错误 .mysql_error（））; 
 $ data = mysql_num_rows（$ result）; 
 $ i = 0; 
 while（$ i< $ data）
 {
 $ s1 = mysql_result（$ result，$ i，  REG_ID）; 
 $ s2 = mysql_result（$ result，$ i，  email_id）; 
 $ s3 = mysql_result（$ result，$ i，  name）; 
 $ s4 = mysql_result（$ result，$ i，  ipd_religion）; 
 $ s5 = mysql_result（$ result，$ i，  ipd_Caste）; 
 $ s6 = mysql_result（$ result，$ i，  ipd_mother_tongue）; 
 $ s7 = mysql_result（$ result，$ i，  ipd_age）; 
 $ s8 = mysql_result（$ result，$ i，  ipd_age1）; 
 $ s9 = mysql_result（$ result，$ i，  ipd_height）; 
 $ s10 = mysql_result（$ result，$ i，  ipd_height1）; 
 $ s13 = mysql_result（$ result，$ i，  ipd_marital_status）; 
 
  $ where  = array（）; 
 
  if （（$ s7！=  ）&&（$ s8！=  ） ）{
 $ where [] =   and age BETWEEN $ s7 AND $ s8; 
} 其他 如果（$ agemin！=   ）{
 $ where [] =  和age> = $ s7; 
} 其他 如果（$ agemax！=   ）{
 $ where [] =  和年龄< = $ s8; 
} 
  if （（$ s9！=  ）&&（$ s10！=   ））{
 $ where [] =   and height BETWEEN $ s9 AND $ s10 ; 
}  else   if （$ hmin！=   ）{
 $ where [] =  和height> = $ s9; 
} 其他 如果（$ hmax！=   ）{
 $ where [] =  和height< = $ s10; 
} 
 
  if （$ s4！=  ）{
 $ where [] =  和religion ='$ s4'; 
} 
 
 如果（$ s13！=  ）{
 $ where [] =  和marital_status ='$ s13'; 
} 
 
 
  $ query1  =  < span class =code-string> SELECT * FROM matri_user_info WHERE gender ='Bride'AND status ='unlock'。加入（$哪里）; 
 $ result1 = mysql_query（$ query1，$ con）或 die（  Mysql错误 .mysql_error（））; 
 $ data1 = mysql_num_rows（$ result1）; 
  echo   $ data1 ; 
 $ i1 = 0; 
 while（$ i1< $ data1）
 {
 $ r1 = mysql_result（$ result，$ i，  REG_ID）; 
 $ r2 = mysql_result（$ result，$ i，  name）; 
 $ r3 = mysql_result（$ result，$ i，  age）; 
 $ r4 = mysql_result（$ result，$ i，  height）; 
 $ r5 = mysql_result（$ result，$ i，  religion）; 
 $ r6 = mysql_result（$ result，$ i，  education）; 
 $ r7 = mysql_result（$ result，$ i，  job）; 
 $ r8 = mysql_result（$ result，$ i，  dob）; 
 $ r9 = mysql_result（$ result，$ i，  candi_pic_1）; 
 
  //  这里我将添加代码以将信息从r1发送到r9。 
 $ i1 ++; 
} 
 $ i ++; 
} 

我想要什么

1- pic一张个人资料表格数据。

2-检查他的偏好并获得data1中的值。

3-在数据1中的所有数据中删除一个配置文件并发送到配置文件1的数据。

4 - 重复循环直到最后。但是每次发生这种情况都应该发送一个独特的个人资料。因为我希望每天只发送一个个人资料。因此，当第二天这个循环将执行来自data1的配置文件1时，应该更改。



如果您有任何想法或建议，请告诉我怎么做。



谢谢

解决方案

query = select * from matri_user_info，其中payment ='是';

result = mysql_query（

查询，

hi,
I am working on a matrimonial website now i have to send a sample profile for every user, i want a unique profile to send to every user but unable to find how to do it.
here is my code what i did.

include("db.php");
$query="select * from matri_user_info where payment='Yes'";
$result=mysql_query($query,$con) or die("Mysql Error".mysql_error());
$data=mysql_num_rows($result);
$i=0;
while($i<$data)
{
$s1=mysql_result($result,$i,"reg_id");
$s2=mysql_result($result,$i,"email_id");
$s3=mysql_result($result,$i,"name");
$s4=mysql_result($result,$i,"ipd_religion");
$s5=mysql_result($result,$i,"ipd_Caste");
$s6=mysql_result($result,$i,"ipd_mother_tongue");
$s7=mysql_result($result,$i,"ipd_age");
$s8=mysql_result($result,$i,"ipd_age1");
$s9=mysql_result($result,$i,"ipd_height");
$s10=mysql_result($result,$i,"ipd_height1");
$s13=mysql_result($result,$i,"ipd_marital_status");

$where = array();
 
if (($s7 !="") && ($s8 !="")) {
	$where[] = " and age BETWEEN $s7 AND $s8";
} else if ($agemin !="") {
	$where[] = " and age >= $s7";
} else if ($agemax!="") {
	$where[] = " and age <= $s8";
}
if (($s9 !="") && ($s10 !="")) {
	$where[] = " and height BETWEEN $s9 AND $s10";
} else if ($hmin !="") {
	$where[] = " and height >= $s9";
} else if ($hmax!="") {
	$where[] = " and height <= $s10";
}
 
if ($s4 !="") {
	$where[] = " and religion = '$s4'";
}

if ($s13 !="") {
	$where[] = " and marital_status = '$s13'";
}
 
 
$query1 = "SELECT * FROM matri_user_info WHERE gender = 'Bride' AND status = 'unlock'" . join($where);
$result1=mysql_query($query1,$con) or die ("Mysql error".mysql_error());
$data1=mysql_num_rows($result1);
echo $data1;
$i1=0;
while($i1<$data1)
{
$r1=mysql_result($result,$i,"reg_id");
$r2=mysql_result($result,$i,"name");
$r3=mysql_result($result,$i,"age");
$r4=mysql_result($result,$i,"height");
$r5=mysql_result($result,$i,"religion");
$r6=mysql_result($result,$i,"education");
$r7=mysql_result($result,$i,"job");
$r8=mysql_result($result,$i,"dob");
$r9=mysql_result($result,$i,"candi_pic_1");

// here i will add code to send information from r1 to r9.
$i1++;
}
$i++;
}

What i want
1- pic one profile form data.
2- Check his preference and get value in data1.
3- Pic one profile among all in data1 and send to profile 1 of data.
4- Repeat loop till last. but every time this happen a unique profile should be sent.because i want only one profile will be send per day. so when next day when this loop will execute for profile 1 value from data1 should be change.

If any one of you have any idea or suggestion then please let me know how to do this.

thanks

解决方案

query="select * from matri_user_info where payment='Yes'";

result=mysql_query(

query,

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