Question

我有来自同一个表的两个查询，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");
}

Answer 1

试试这个（它不是你问题的完整解决方案，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;

};

了解如何使用预备声明!!!! http://php.net/manual/en/mysqli-stmt.bind-param.php

Answer 2

<强> 1。您要分配而不是比较：

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
}

注意单=个？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
}

由于这个原因，我更喜欢'' == $name而不是$name == ''因为'' = $name会导致解析错误，这比正在运行但已损坏的脚本更好。

<强> 2。您没有将$row传递给find_pic()：

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

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);

第3。 $count1和$count2未定义：

您正在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);

Answer 3

如果要获得两张随机图片，为什么不这样做？

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

具有不同结果名称的多个mysql查询

3 个答案: