更改JSON的mysql结果以按键索引

Question

I have a mysql result that I'm fetching as an array and encoding to JSON like so:

$getDisplayPage = "
SELECT p.id as pageID, page_type_id, display_id, slide_order, duration, background_img, pn.ID as panel_id, panel_type_id, cont_id, c.ID as contID, content 
FROM pages p
inner join panels pn
on p.id = pn.page_id 
inner join content c
on pn.cont_id = c.id
WHERE p.active = 1
and pn.active = 1
AND p.display_id = '".$display."'
";

$showDisplayResult = $mysqlConn->query($getDisplayPage);
while($row=mysqli_fetch_assoc($showDisplayResult))
    {
        $rows[] = $row;
    }
$showDisplays = json_encode($rows);

Which returns a row for each entry, but I need to somehow change it to index by page ID. If you see my JSON below, it returns the correct data but it should actually only have 3 objects/rows as opposed to 5. I need to be able to access all attributes of a given pageID. How can I alter this to give me the correct JSON object?

[{"pageID":"93",
    "page_type_id":"2",
    "display_id":"2",
    "slide_order":null,
    "duration":"74",
    "background_img":"images\/bg_rainbow.svg",
    "panel_id":"86",
    "panel_type_id":"2",
    "cont_id":"138",
    "contID":"138",
    "content":"


<\/head>

Left 93<\/p>
<\/body>
<\/html>"},
{"pageID":"93",
    "page_type_id":"2",
    "display_id":"2",
    "slide_order":null,
    "duration":"74",
    "background_img":"images\/bg_rainbow.svg",
    "panel_id":"87",
    "panel_type_id":"3",
    "cont_id":"139",
    "contID":"139",
    "content":"


<\/head>

Right 93<\/p>
<\/body>
<\/html>"},
{"pageID":"95",
    "page_type_id":"1",
    "display_id":"2",
    "slide_order":null,
    "duration":"123",
    "background_img":"images\/bg_rainbow.svg",
    "panel_id":"90",
    "panel_type_id":"1",
    "cont_id":"142",
    "contID":"142",
    "content":"


<\/head>

Testing a full page for ID 95<\/p>
<\/body>
<\/html>"},
{"pageID":"105",
    "page_type_id":"2",
    "display_id":"2",
    "slide_order":null,
    "duration":"54",
    "background_img":"images\/bg_rainbow.svg",
    "panel_id":"97",
    "panel_type_id":"2",
    "cont_id":"149",
    "contID":"149",
    "content":"


<\/head>

This is left content<\/p>
<\/body>
<\/html>"},
{"pageID":"105",
    "page_type_id":"2",
    "display_id":"2",
    "slide_order":null,
    "duration":"54",
    "background_img":"images\/bg_rainbow.svg",
    "panel_id":"98",
    "panel_type_id":"3",
    "cont_id":"150",
    "contID":"150",
    "content":"


<\/head>

This is right content<\/p>
<\/body>
<\/html>"}]

Answer 1

$getDisplayPage = "
SELECT p.id as pageID, page_type_id, display_id, slide_order, duration, background_img, pn.ID as panel_id, panel_type_id, cont_id, c.ID as contID, content 
FROM pages p
inner join panels pn
on p.id = pn.page_id 
inner join content c
on pn.cont_id = c.id
WHERE p.active = 1
and pn.active = 1
AND p.display_id = '".$display."'
";

$showDisplayResult = $mysqlConn->query($getDisplayPage);
while($row=mysqli_fetch_assoc($showDisplayResult))
    {
        $rows[$row['pageID']][] = $row;
    }
$showDisplays = json_encode($rows);