Question

我有条形码扫描仪应用程序。我希望在Android应用程序本身上扫描条形码，但结果/条形码编号将重定向到新的php页面。任何人都知道如何做到这一点？下面的代码是我们扫描条形码的编码：

public class CameraTestActivity extends Activity
 {
private Camera mCamera;
private CameraPreview mPreview;
private Handler autoFocusHandler;
URL url;


TextView scanText;
Button scanButton;

ImageScanner scanner;

private boolean barcodeScanned = false;
private boolean previewing = true;

static {
    System.loadLibrary("iconv");
} 

public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);

    setContentView(R.layout.activity_scan);

    setRequestedOrientation(ActivityInfo.SCREEN_ORIENTATION_PORTRAIT);

    autoFocusHandler = new Handler();
    mCamera = getCameraInstance();

    /* Instance barcode scanner */
    scanner = new ImageScanner();
    scanner.setConfig(0, Config.X_DENSITY, 3);
    scanner.setConfig(0, Config.Y_DENSITY, 3);

    mPreview = new CameraPreview(this, mCamera, previewCb, autoFocusCB);
    FrameLayout preview = (FrameLayout)findViewById(R.id.cameraPreview);
    preview.addView(mPreview);

    scanText = (TextView)findViewById(R.id.scanText);

    scanButton = (Button)findViewById(R.id.ScanButton);

    scanButton.setOnClickListener(new OnClickListener() {
            public void onClick(View v) {
                if (barcodeScanned) {
                    barcodeScanned = false;
                    scanText.setText("Scanning...");
                    mCamera.setPreviewCallback(previewCb);
                    mCamera.startPreview();
                    previewing = true;
                    mCamera.autoFocus(autoFocusCB);
                    Intent viewIntent =
                            new Intent("android.intent.action.VIEW",
                                        Uri.parse("http://172.20.138.136/searchform.php"));
                            startActivity(viewIntent);
                }
            }
        });

}

public void postdata()
{

}

public void onPause() {
    super.onPause();
    releaseCamera();
}


/** A safe way to get an instance of the Camera object. */
public static Camera getCameraInstance(){
    Camera c = null;
    try {
        c = Camera.open();

这是我们的PHP代码：

<html>
<body>
<form action="insertform.php" method="post">
 &nbsp; <input type="text" name="t1" /><br />
 <input type="submit" value="Search" />
</form>

</body>



 </html>

Answer 1

您可以使用此功能通过意图打开您的页面：

 Intent i = new Intent(Intent.ACTION_VIEW, 
           Uri.parse("http://172.20.138.136/searchform.php"));
    startActivity(i);

条形码扫描仪结果重定向到新的PHP页面

1 个答案: