将Mysql查询结果信息放在多维数组中

Question

#更新了这个表:明细表

我想把结果放在这个json数组中：

{
    "selectedColor": "black",
      "black": {
        "thumb": {
          "image1": "../img/e-commerce/product/dark-small-1.jpg",
          "image2": "../img/e-commerce/product/dark-small-2.jpg",
          "image3": "../img/e-commerce/product/dark-small-3.jpg"
        },
        "large": {
          "image1": "../img/e-commerce/product/dark-large-1.jpg",
          "image2": "../img/e-commerce/product/dark-large-2.jpg",
          "image3": "../img/e-commerce/product/dark-large-3.jpg"
        }
      },
    "selectedSlide": 0
  }

到目前为止，在@IT高盛的帮助下，我一直在努力做到这一点：

$sImg = "SELECT prod_thumb160x90, prod_img1280x720 FROM product_img_detail WHERE prod_id=".$data2['id'];
$qPrd = mysql_query($sImg);
$rPrd = mysql_fetch_array($qPrd);

$imgPath = "img/e-commerce/product/";

$count = 0;
$arrSm = [];
$arrLg = [];

foreach ($rPrd as $row) {
    $count++;
    $arrSm["image$count"] = $imgPath.$rPrd['prod_thumb160x90'];
    $arrLg["image$count"] = $imgPath.$rPrd['prod_img1280x720'];
}

$arrImg = array("thumb"=>$arrSm, "large"=>$arrLg);
$res = array("selectedColor"=>"black", "black"=>$arrImg, "selectedSlide"=>0);

echo $json = json_encode($res);

上述代码的输出如下：

  {
"selectedColor": "black",
"black": {
    "thumb": {
        "image1": "img\/e-commerce\/product\/dark-small-1.jpg",
        "image2": "img\/e-commerce\/product\/dark-small-1.jpg",
        "image3": "img\/e-commerce\/product\/dark-small-1.jpg",
        "image4": "img\/e-commerce\/product\/dark-small-1.jpg"
    },
    "large": {
        "image1": "img\/e-commerce\/product\/dark-large-1.jpg",
        "image2": "img\/e-commerce\/product\/dark-large-1.jpg",
        "image3": "img\/e-commerce\/product\/dark-large-1.jpg",
        "image4": "img\/e-commerce\/product\/dark-large-1.jpg"
    }
},
"selectedSlide": 0

}

我很好奇，如果我将查询更改为从表中选择*，它一直循环到"image10"，如果我将查询更改为只从表中选择1列，则在"image2“处停止循环！

有人能帮我吗？请帮助我用正确有效的代码进行重组(并解开正反斜杠)，我已经困了3天了.

Answer 1

在与@IT高盛代码结合之后，我找到了答案：

$sImg = "SELECT * FROM product_img_detail WHERE prod_id=".$data2['id'];
$qPrd = mysql_query($sImg);

$imgPath = "img/e-commerce/product/";

$count = 0;
$arrSm = [];
$arrLg = [];

//foreach ($rPrd as $row) {
while($rPrd=mysql_fetch_array($qPrd)){
    $count++;
    $arrSm["image$count"] = $imgPath.$rPrd['prod_thumb160x90'];
    $arrLg["image$count"] = $imgPath.$rPrd['prod_img1280x720'];
}

$arrImg = array("thumb"=>$arrSm, "large"=>$arrLg);
$res = array("selectedColor"=>"black", "black"=>$arrImg, "selectedSlide"=>0);

echo $json = json_encode($res);

Answer 2

您需要循环记录(行)并使用正确的键(根据$count)聚合到数组，您需要循环记录(行)，并使用正确的键(根据$count)聚合到数组中。

编辑:我修正了现在不再推荐的mysql_fetch_array的用法

<?php

$sql = "SELECT * FROM product_img_detail WHERE prod_id = " . $data2['id'];
$qPrd = mysql_query($sql);
$imgPath = "../img/e-commerce/product/";

$count = 0;
$arrSm = [];
$arrLg = [];

while ($row = mysql_fetch_array($qPrd, MYSQL_ASSOC)) {
    $count++;
    $arrSm["image$count"] = $imgPath . $row['prod_thumb160x90'];
    $arrLg["image$count"] = $imgPath . $row['prod_img1280x720'];
}

mysql_free_result($qPrd);

$arrImg = array("thumb" => $arrSm, "large" => $arrLg);
$res = array("selectedColor" => "black", "black" => $arrImg, "selectedSlide" => 0);

echo $json = json_encode($res);