使用 json 选择多个 mysqli答案

【问题标题】：Multiple mysqli select with json使用 json 选择多个 mysqli
【发布时间】：2014-08-03 15:08:27
【问题描述】：

<?php
include"../database_conn.php";
$con=mysqli_connect("localhost","admin","123456","ayurveadic");

$query_pag_data = "SELECT id from diseases";
$result_pag_data = mysql_query($query_pag_data) or die('MySql Error' . mysql_error());

while ($row = mysql_fetch_array($result_pag_data)) {
    $diseases_id = $row['id'];

    $result = mysqli_query($con,"SELECT $diseases_id FROM treatment WHERE gender_id = '10' AND diseases_id = '$diseases_id'");
    $row_gid = mysqli_fetch_array($result);
    if ($row_gid == TRUE){

        $sl_dise = mysqli_query($con,"SELECT Diseases_type, id FROM diseases WHERE id = '$diseases_id'");

        $rowss = array();
        while($r = mysqli_fetch_assoc($sl_dise)) {
            $rowss[] = $r;
        }
        print json_encode($rowss);
    }
}

输出是：

[{"Diseases_type":"fever","id":"114"}][{"Diseases_type":"rhrh","id":"123"}]

我怎样才能得到这个输出：

[{"Diseases_type":"fever","id":"114"},{"Diseases_type":"rhrh","id":"123"}]

【问题讨论】：

为什么要混合 mysql_ 和 mysqli_。总是首选 mysqli_，因为 mysql_ 已被弃用。
PHP combine arrays into one big array的可能重复

标签： php mysql json select

【解决方案1】：

把你的

print json_encode($rowss);

结束后while ($row = mysql_fetch_array($result_pag_data)) {

也去掉这个$rowss = array();数组初始化

【讨论】：

这不是他需要改变的全部，因为他每次循环都在重新初始化$rowss。

【解决方案2】：

这是你优化的代码

<?php
include"../database_conn.php";
$con=mysqli_connect("localhost","admin","123456","ayurveadic");

 $query_pag_data = "select diseases.Diseases_type,diseases.id from diseases inner join treatment on treatment.diseases_id = diseases.id where treatment.gender_id = '10' ";
$result_pag_data = mysql_query($query_pag_data) or die('MySql Error' . mysql_error());

$rowss = array();
while ($row = mysqli_fetch_assoc($result_pag_data)) {
    $rowss[] = $row;
}
print json_encode($rowss);
?>

【讨论】：

为什么将SELECT Diseases_type, id 替换为SELECT *？我们应该鼓励人们选择特定的列。
@Barmar 好的。同意并更新代码。但为什么要投反对票？
我没有投反对票。我最初投了赞成票，然后撤消了它，直到你解决了这个问题。

【解决方案3】：

您需要在外循环之前初始化 JSON 数组，并在最后打印它，而不是每次都打印。

<?php
include"../database_conn.php";
$con=mysqli_connect("localhost","admin","123456","ayurveadic");

$query_pag_data = "SELECT id from diseases";
$result_pag_data = mysql_query($query_pag_data) or die('MySql Error' . mysql_error());

$rowss = array();
while ($row = mysql_fetch_array($result_pag_data)) {
    $diseases_id = $row['id'];

    $result = mysqli_query($con,"SELECT $diseases_id FROM treatment WHERE gender_id = '10' AND diseases_id = '$diseases_id'");
    $row_gid = mysqli_fetch_array($result);
    if ($row_gid == TRUE){

        $sl_dise = mysqli_query($con,"SELECT Diseases_type, id FROM diseases WHERE id = '$diseases_id'");

        while($r = mysqli_fetch_assoc($sl_dise)) {
            $rowss[] = $r;
        }
    }
}
print json_encode($rowss);

【讨论】：