我需要将mysql查询的结果附加到另一个查询的现有结果中,这两个查询都是在不同的表上执行的,实际上,我真正想做的是从主表中获取值,并使用第一个值从另一个表中获取一些数据,然后使用json编码将所有这些数据组合在一起。
$query = "select ID,TIMESTAMP,UUID from TABLE1 where USERNAME='$user'";
$result = mysqli_query($conn, $query);
$numrows = mysqli_num_rows($result);
if($numrows>0)
{
$res=$a;
$myArray = array();
while($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
$eachuuid = $row[UUID];
$query1 = "select ID,IMAGEURL from TABLE_IMAGES where IMG_UUID='$eachuuid'";
$result1 = mysqli_query($conn, $query1);
$numrows1 = mysqli_num_rows($result);
if($numrows1>0){
$res1;
$myArray1 = array();
while($row1 = mysqli_fetch_array($result1, MYSQLI_ASSOC)) {
$myArray1[] = $row1;
}
$row['IMG_URL']=$myArray1;
}
$myArray[] = $row;
}
$res['Data']=$myArray;
echo json_encode($res);结果:
{"Response":"OK","Data":[{"ID":"62",
"TIMESTAMP":"26 January",
"UUID":"12345",
"IMG_URL":[{"ID":"5","IMAGEURL":"26_January_2016_22_39_28_crop.jpg"}]}]}我需要得到输出,
{"Response":"OK","Data":[{"ID":"62",
"TIMESTAMP":"26 January",
"UUID":"12345",
"IMG_URL":{"ID":"5","IMAGEURL":"26_January_2016_22_39_28_crop.jpg"}}]}意味着需要从IMG_URL部分删除[],这样我就可以轻松地解析数据。
发布于 2016-01-27 06:12:41
你能做到的
$row['IMG_URL']=$myArray1[0];或
$myArray1 = null;
while($row1 = mysqli_fetch_array($result1, MYSQLI_ASSOC)) {
$myArray1 = $row1;
}
$row['IMG_URL']=$myArray1;我希望这能帮到你。
https://stackoverflow.com/questions/35029995
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