如何执行此查询？

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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 = 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,