duanna3634 2016-08-02 05:54
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如何将查询数据放在两个不同的数组中然后在两个表中回显?

Run select query on table

filter the query data

if its gender equal to Man, then put all of his in array men=array()

OR

if its gender is woman then put in array woman=array()

then echo two tables

Men Table

1-Name, age, city, country, etc

2-Name, age, city, country, etc

3-Name, age, city, country, etc . . .

Women Table

1-Name, age, city, country, etc

2-Name, age, city, country, etc

3-Name, age, city, country, etc . . . .

Men table will contain all rows data of Man which are save in Men() array

Women table will contain all rows data of Woman which are save in Women() array

Note:-

I have tried code like this but its mess and i failed:-

https://stackoverflow.com/questions/38653118/notice-undefined-offset-1-putting-mysql-query-data-in-while-loop

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  • dongyuduan1890 2016-08-02 06:24
    关注

    your query must be in UNION:

    Select men.Name, men.age, men.city, men.country, 'Male' as 'gender' FROM men_table as men UNION ALL Select women.Name, women.age, women.city, women.country, 'Female' as 'gender FROM women_table as women;
    

    so you can have an query output:

    |Name   |age    |city   |country|gender|
    ----------------------------------------
    |male1  |21     |NY     |USA    |Male  |
    |male2  |23     |CLV    |USA    |Male  |
    |female1|25     |GS     |USA    |Female|
    |female2|27     |CHG    |USA    |Female|
    

    after that you can loop to your result:

    <?php 
    
    $mysqli = new mysqli("localhost", "my_user", "my_password", "world");
    
    /* check connection */
    if ($mysqli->connect_errno) {
        printf("Connect failed: %s
    ", $mysqli->connect_error);
        exit();
    }
        $query = "Select men.Name, men.age, men.city, men.country, 'Male' as 'gender' FROM men_table as men UNION ALL Select women.Name, women.age, women.city, women.country, 'Female' as 'gender FROM women_table as women";
    
        $men = array();
        $women = array;
    
        if ($result = $mysqli->query($query)) {
    
            /* fetch associative array */
            while ($row = $result->fetch_assoc()) {
                if($row["gender"] == 'Male') {
                    $men[] = $row[];
                }
                else {
                    $women[] = $row[];
                }
            }
    
            print_r($men);
            print_r($women);
    
            /* free result set */
            $result->free();
        }
    
    /* close connection */
    $mysqli->close();
    ?>
    

    The output will be like:

    //$men array
        array(
            [0] => array(
                    'Name' => 'male1', 
                    'age' => '21', 
                    'city' => 'NY', 
                    'country' => 'USA', 
                    'gender' => 'Male'
            ),
            [1] => array(
                    'Name' => 'male2', 
                    'age' => '23', 
                    'city' => 'CLV', 
                    'country' => 'USA', 
                    'gender' => 'Male'
            )
        );
    
    //$women array
        array(
            [0] => array(
                    'Name' => 'female1', 
                    'age' => '25', 
                    'city' => 'GS', 
                    'country' => 'USA', 
                    'gender' => 'Female'
            ),
            [1] => array(
                    'Name' => 'female2', 
                    'age' => '27', 
                    'city' => 'CHG', 
                    'country' => 'USA', 
                    'gender' => 'Female'
            )
        );
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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