dongmuyuan3046 2017-02-23 09:42
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Mysql或PHP跳过第一行数据库[关闭]

I have been trying to make a php function that uses mysql to output three columns of records into a 2-D list. However when doing this, the first value in the database is skipped every time it has been run. I have read through the code again, and again and I cant seem to find any problems. The problem is not the output loop at the end as no matter where I put the echo, the first record in the database will not show up. I am using XAMPP to run the code by the way. In chrome browser.

<?php

function countHouse() {
    $connection = mysql_connect('localhost','','');
    mysql_select_db('sports');

    $query = "SELECT ID, house, housePoints FROM sportsday WHERE housePoints     != 0";
    $result = mysql_query($query);
    $row = mysql_fetch_array($result);
    $data = array();
    while($row = mysql_fetch_array($result)) {
        $data[] = array($row['house'],$row['housePoints'],$row['ID']);
    }
    mysql_close();
    return $data;
}

//Output, not a problem except $data is missing first term
$data = countHouse();
for ($i=0; $i<count($data); $i++) {
    echo $data[$i][2] . " " .$data[$i][0] . $data[$i][1];?><html><br></html>     <?php
}
?>
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1条回答 默认 最新

  • dongpinken0498 2017-02-23 09:46
    关注

    Notice that you are reading the first row before the while:

    // get rid of this line of code
    //$row = mysql_fetch_array($result);
    $data = array();
    while($row = mysql_fetch_array($result)) {
        $data[] = array($row['house'],$row['housePoints'],$row['ID']);
    }
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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