具有不同结果名称的多个mysql查询

I have two queries from the same table and selecting the same info but sending results to two names. There are two variables to store the result, but how do I store other information from the queries?

我从同一个表中有两个查询并选择相同的信息,但将结果发送到两个名称。存储结果有两个变量,但如何存储查询中的其他信息?

require_once('connectvars.php');
$dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME) or die('Error connecting to server, line 20.');
$query = "SELECT COUNT(id)  FROM artwork"; //find total number of id's
$result = mysqli_query ($dbc, $query) or die("query error, line 22");
$row = mysqli_fetch_array ($result, MYSQL_NUM);
find_pic(); //get the two id numbers
while ($count1 = $count2 or name = ""){ //problem line. check if counts = each other or either name = nothing
    find_pic(); //if true, run function again
}
while ($count1=37 or $count2=37){ //if either count = 37
    find_pic(); //if true, run function again
}
show_pic();

function find_pic(){
    $count1 = rand(0,$row[0]);
    $count2 = rand(0,$row[0]);
    $query1 = "SELECT * FROM artwork WHERE id = $count1";
    $query2 = "SELECT * FROM artwork WHERE id = $count2";
    $result1 = mysqli_query ($dbc, $query1) or die("query error, line 38");
    $result2 = mysqli_query ($dbc, $query2) or die("query error, line 39");
}

3 个解决方案

#1

If the whole point is to get two random pictures why not do this?

如果要获得两张随机图片,为什么不这样做呢?

SELECT
    *
FROM
    artwork
WHERE 
    id != 37 AND
    name != ''
ORDER BY 
    RAND()
LIMIT 2

#2

Try this (it isnt complete solution for your problem, its only help):

试试这个(它不是你的问题的完整解决方案,它唯一的帮助):

$result1 = $dbc->prepare("SELECT * FROM artwork WHERE id = $count1");
$result2 = $dbc->prepare("SELECT * FROM artwork WHERE id = $count2");
$result1->execute();
$result2->execute();
$result1->bind_result($column1, $column2);   //all columns from table artwork
$result2->bind_result($column1, $column2);   //...


while ($result1->fetch())
{
    echo $column1;                       //here u have your result data
        echo $column2;

};

Learn how to use PREPARED STATEMENTS!!!! http://php.net/manual/en/mysqli-stmt.bind-param.php

学习如何使用准备好的声明!!!! http://php.net/manual/en/mysqli-stmt.bind-param.php

#3

1. You're assigning rather than comparing:

你在分配而不是比较:

while ($count1 = $count2 or name = ""){ //problem line. check if counts = each other or either name = nothing
    find_pic(); //if true, run function again
}
while ($count1=37 or $count2=37){ //if either count = 37
    find_pic(); //if true, run function again
}

Notice the single =s? You want two (or three):

注意单个= s?你想要两个(或三个):

while ($count1 == $count2 or name == ""){ //problem line. check if counts = each other or either name = nothing
    find_pic(); //if true, run function again
}
while ($count1==37 or $count2==37){ //if either count = 37
    find_pic(); //if true, run function again
}

It's for this reason that I prefer to do '' == $name rather than $name == '' as '' = $name will result in a parse error which is better than a running but broken script.

出于这个原因,我更喜欢做''== $ name而不是$ name ==''作为''= $ name将导致解析错误,这比运行但破坏的脚本更好。

2. You're not passing $row to find_pic():

2.你没有将$ row传递给find_pic():

Most variables are subject to scope. $row doesn't exist inside the find_pic function unless you pass it as a parameter:

大多数变量都受范围限制。除非您将其作为参数传递,否则find_pic函数中不存在$ row:

function find_pic($row){
    $count1 = rand(0,$row[0]);
    $count2 = rand(0,$row[0]);
    $query1 = "SELECT * FROM artwork WHERE id = $count1";
    $query2 = "SELECT * FROM artwork WHERE id = $count2";
    $result1 = mysqli_query ($dbc, $query1) or die("query error, line 38");
    $result2 = mysqli_query ($dbc, $query2) or die("query error, line 39");
}

find_pic($row);

3. $count1 and $count2 are not defined:

3. $ count1和$ count2未定义:

You're setting them inside the find_pic function but they don't exist outside that scope. You might want to use references:

您在find_pic函数中设置它们,但它们不在该范围之外。您可能想要使用引用:

function find_pic($row, &$count1, &$count2){
    $count1 = rand(0,$row[0]);
    $count2 = rand(0,$row[0]);
    $query1 = "SELECT * FROM artwork WHERE id = $count1";
    $query2 = "SELECT * FROM artwork WHERE id = $count2";
    $result1 = mysqli_query ($dbc, $query1) or die("query error, line 38");
    $result2 = mysqli_query ($dbc, $query2) or die("query error, line 39");
}

find_pic($row, $count1, $count2);

#1