在我的 Activity A 中，它有一个 listView，其中的图像和文本是从 Activity B 获取的。
Activtiy 带有图像和文本的 ListView
当点击保存按钮时，我想通过php将图片路径和文本存入mysql，图片将存储在目录中。
Activity A
public void uploadImageAndText(ArrayList<ImageAndText> listItems, final String id) {
JSONArray jsonArray = new JSONArray();
try {
for (ImageAndText i : listItems) {
JSONObject object = new JSONObject();
String type = i.getType();
String[] Type = type.split(":");
object.put("type", Type[1]);
Toast.makeText(getApplicationContext(), Type[1], Toast.LENGTH_LONG).show();
String amount = i.getAmount();
String[] Amount = amount.split(":");
object.put("amount", Amount[1]);
String description = i.getDescription();
String[] Description = description.split(":");
object.put("description", Description[1]);
Bitmap uploadImage = i.getImage();
String image = getStringImage(uploadImage);
object.put("image", image);
object.put("ts_id", id);
jsonArray.put(object);
}
} catch (JSONException e) {
e.printStackTrace();
}
AddStaff ru = new AddStaff(jsonArray);
ru.execute();
}
class AddStaff extends AsyncTask<String, Void, String> {
ProgressDialog loading;
JSONArray jsonArray;
AddStaff(JSONArray jsonArray) {
this.jsonArray = jsonArray;
}
@Override
protected void onPreExecute() {
super.onPreExecute();
loading = ProgressDialog.show(AddClaims.this, "Please Wait", null, true, true);
}
@Override
protected String doInBackground(String... params) {
HashMap<String, String> data = new HashMap<String, String>();
data.put("listItems", jsonArray.toString());
RequestHandler rh = new RequestHandler();
String result = rh.sendPostRequest(Configs.STAFF_BENEFIT, data);
return result;
}
@Override
protected void onPostExecute(String s) {
super.onPostExecute(s);
loading.dismiss();
Toast.makeText(getApplicationContext(), s, Toast.LENGTH_LONG).show();
}
}
public String getStringImage(Bitmap bmp) {
ByteArrayOutputStream baos = new ByteArrayOutputStream();
bmp.compress(Bitmap.CompressFormat.JPEG, 100, baos);
byte[] imageBytes = baos.toByteArray();
String encodedImage = Base64.encodeToString(imageBytes, Base64.DEFAULT);
return encodedImage;
}
}
php
<?php
if( $_SERVER['REQUEST_METHOD']=='POST' ){
if( !empty( $_POST['listItems'] ) ){
$mysqli = new mysqli("127.0.0.1:3307", "root", "", "androiddb");
if( $mysqli->connect_errno ) echo "Failed to connect to MySQL";/* try not to reveal too much info! */
/* The example you followed had this line! */
$image = $_POST['image'];
$listItems = json_decode( $_POST['listItems'], true );
/* This never gets called - should it? */
$sql="SELECT id FROM staff_benefit ORDER BY id ASC";
/*
This fetches ALL ids from the staff_benefit table
so which ID do you need? Iterating through the
recordset will effectively return the last one!
*/
$id=0;
/*
I assume that $id is supposed to get
the value from th above sql query???
*/
$res=$mysqli->query( $sql );
while( $rs=$res->fetch_object() ) $id=$rs->id;
$path="$id.png";
$actualpath="http://192.168.107.115:80/Android/CRUD/PhotoUpload/$path";
/* Use only placeholders in your prepared statement */
$sql="INSERT INTO `staff_benefit` ( `type`, `amount`, `description`, `image`, `ts_id` ) VALUES ( ?, ?, ?, ?, ? )";
$stmt=$mysqli->prepare( $sql );
/* Save the image to disk: I prefer to use the actual path for storing files */
$pathelements=array( realpath( $_SERVER['DOCUMENT_ROOT'] ), 'CRUD', 'PhotoUpload', '' );/* empty entry adds trailing slash to path */
$savepath = realpath( implode( DIRECTORY_SEPARATOR, $pathelements ) ) . "{$id}.png";
$bytes=file_put_contents( $savepath, base64_decode( $image ) );
if( !$bytes ){
echo 'Error saving image';
}
if ( $stmt && $bytes ) {
foreach( $listItems as $item ){
$stmt->bind_param('sssss', $item['type'], $item['amount'], $item['description'], $actualpath, $item['ts_id'] );
$res=$stmt->execute();
if( !$res ) echo 'Query failed with code: '.$stmt->errno;/* Again, not too much info */
}
}
$mysqli->close();
}
}
?>
When click image in directory, can't display image because the file is empty
最佳答案
<?php
if( $_SERVER['REQUEST_METHOD']=='POST' ){
if( !empty( $_POST['listItems'] ) ){
$mysqli = new mysqli("127.0.0.1:3307", "root", "", "androiddb");
if( $mysqli->connect_errno ) echo "Failed to connect to MySQL";/* try not to reveal too much info! */
/* The example you followed had this line! */
$image = $_POST['image'];
$listItems = json_decode( $_POST['listItems'], true );
/* This never gets called - should it? */
$sql="SELECT id FROM staff_benefit ORDER BY id ASC";
/*
This fetches ALL ids from the staff_benefit table
so which ID do you need? Iterating through the
recordset will effectively return the last one!
*/
$id=0;
/*
I assume that $id is supposed to get
the value from th above sql query???
*/
$res=$mysqli->query( $sql );
while( $rs=$res->fetch_object() ) $id=$rs->id;
$path="$id.png";
$actualpath="http://192.168.107.115:80/Android/CRUD/PhotoUpload/$path";
/* Use only placeholders in your prepared statement */
$sql="INSERT INTO `staff_benefit` ( `type`, `amount`, `description`, `image`, `ts_id` ) VALUES ( ?, ?, ?, ?, ? )";
$stmt=$mysqli->prepare( $sql );
/* Save the image to disk: I prefer to use the actual path for storing files */
$pathelements=array( realpath( $_SERVER['DOCUMENT_ROOT'] ), 'CRUD', 'PhotoUpload', '' );/* empty entry adds trailing slash to path */
$savepath = realpath( implode( DIRECTORY_SEPARATOR, $pathelements ) ) . "{$id}.png";
$bytes=file_put_contents( $savepath, base64_decode( $image ) );
if( !$bytes ){
echo 'Error saving image';
}
if ( $stmt && $bytes ) {
foreach( $listItems as $item ){
$stmt->bind_param('sssss', $item['type'], $item['amount'], $item['description'], $actualpath, $item['ts_id'] );
$res=$stmt->execute();
if( !$res ) echo 'Query failed with code: '.$stmt->errno;/* Again, not too much info */
}
}
$mysqli->close();
}
}
?>
关于php - 想要将图像路径存储到 MySQL 时遇到麻烦，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/34717255/