我想从php中的json中检索图像。实际上，我想通过php将这些图像源作为json数据获取，并使用javascript进行显示。但我觉得我做错了什么，请帮帮我。
<script type="text/javascript">
function jsonGetImages(name){
var thumbnailbox = document.getElementById("picturebox");
var hr = new XMLHttpRequest();
hr.open("POST", "jsonget.php", true);
hr.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
hr.onreadystatechange = function() {
if(hr.readyState == 4 && hr.status == 200) {
var d = JSON.parse(hr.responseText);
picturebox.innerHTML = "";
for(var o in d){
if(d[o].src){
picturebox.innerHTML += '<img src="'+d[o].src+'">';
}
}
}
}
hr.send("name="+name);
picturebox.innerHTML = "requesting...";
}
</script>
</head>
<body>
<div id="picturebox"></div>
<script type="text/javascript">jsonGetImages('Jason');</script>
</body>
jsonget.php
<?php
header("Content-Type: application/json");
$folder = 'images';
$cn=mysql_connect("localhost","root","");
if(!$cn){
echo "<b>Connection couldn't be established.</b>";
die();
}
$db=mysql_select_db("test",$cn);
if(!$db){
echo "<b>Database doesn't exist.</b>";
}
$dir = $folder."/";
$dirHandle = opendir($dir);
$name=$_POST['name'];
$sql="SELECT * FROM users WHERE name='$name'";
$result=mysql_query($sql);
echo mysql_num_rows($result);
$i=0;
$jsonData = '{';
while($row=mysql_fetch_array($result)){
$name=$row['name'];
$image_name=$row['image'];
$i++;
$file=readdir($dirHandle);
$src = "$dir$image_name";
$jsonData .= '"img'.$i.'":{ "num":"'.$i.'","src":"'.$src.'", "name":"'.$name.'" },';
}
closedir($dirHandle);
$jsonData = chop($jsonData, ",");
$jsonData .= '}';
echo $jsonData;
?>
请帮忙。我找不到如何使用json获取图像的解决方案。
您可以使用php将图像转换为Base64字符串，并使用JSON 进行传递
$src = "$dir$image_name";
$type = pathinfo($src, PATHINFO_EXTENSION);
$data = file_get_contents($src);
$base64 = 'data:image/' . $type . ';base64,' . base64_encode($data);
$myobj = json_encode(array("image"=>$base64));
然后，您可以简单地使用javascript将该字符串放入图像元素的SRC属性中。
var obj = JSON.parse(serverResponse);
img.src=obj.image;
$res = mysqli_query($conn,"SELECT * FROM table_name");
while($row = mysqli_fetch_array($res)){
echo '<img src="data:image/jpeg;base64,'.base64_encode($row['al_image']).'" >';
}
我使用上面的代码来显示图像。它是加密的。之后，您可以简单地通过JS中的src来显示图像。
我使用此代码返回保存在MySQL数据库中的文本和图像。我使用PHP进行访问，返回的数据运行在Android应用程序上。
<?php
/*******************************************************
{ }
{ GET IMAGE JSON }
{ }
{ File: GetImageJson.php }
{ Copyright (c) Zicatti Software 2015 }
{ Developer: Osmir Zicatti }
{ }
{ }
{ Used to return the fields in a table }
{ containing image ( BLOB ) using json }
{ }
{*******************************************************}
{*******************************************************}
{ Paramentro necessário: Numero da pagina }
{ Formato de chamada: GetImageJson.php?PAGINA=1 }
{*******************************************************/
$Pagina = @$_GET['PAGINA'];
if (@$_GET['PAGINA'] != '') {
$host="200.200.200.200"; // Host name
$username="BDados"; // Mysql username
$password="password"; // Mysql password
$db_name="DATABASE"; // Database name
// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
mysql_set_charset('utf8');
// retorna TODOS os campos com as 2 imagens
// $qry = sprintf("SELECT id, pagina, titulo, texto1, texto2, image_thumb, imagem FROM TABELA where pagina = %s",
// Mas vou usar esta para carregar a ListView com o thunmbnail que é menor
$qry = sprintf("SELECT id, pagina, titulo, texto1, texto2, image_thumb FROM TABELA where pagina = %s",
mysql_real_escape_string($Pagina));
$query=mysql_query($qry);
if (!$query) {
$message = 'Invalid query: ' . mysql_error() . "'n";
$message .= 'Whole query: ' . $qry;
die($message);
}
$return_arr = array();
$row_array = array();
// Verifica se existe algum registro
$num_rows = mysql_num_rows($query);
if ($num_rows > 0) {
while ($r = mysql_fetch_array($query)) {
$row_array['id'] = $r['id'];
$row_array['pagina'] = $r['pagina'];
$row_array['titulo'] = $r['titulo'];
$row_array['texto1'] = $r['texto1'];
$row_array['texto2'] = $r['texto2'];
$row_array['image_thumb'] = base64_encode($r['image_thumb']);
array_push($return_arr,$row_array);
}
}
else { $return_arr['id'] = 'ERRO - Pagina inexistente'; }
echo json_encode($return_arr);
return json_encode($return_arr);
mysql_close();
}
else {
$return_arr['id'] = 'ERRO - Faltou enviar o numero da pagina';
echo json_encode($return_arr);
return json_encode($return_arr);
}
?>