我正在建立一个订单打印应用程序，我最初遇到了一个问题，我自己解决了。我需要在checkListArrayList中设置标签位置，以便无论商品名称有多长，价格都会排成一行。我已经遗漏了下面的声明，因为它们与我的问题无关，但我这样做是使用：
Process: com.example.siddhi.meavita, PID: 20503
java.lang.IndexOutOfBoundsException: Invalid index 5, size is 5
at java.util.ArrayList.throwIndexOutOfBoundsException(ArrayList.java:255)
at java.util.ArrayList.get(ArrayList.java:308)
at com.example.siddhi.meavita.Activities.CheckListActivity$GetCheckListsItemAsyncTask.onPostExecute(CheckListActivity.java:336)
at com.example.siddhi.meavita.Activities.CheckListActivity$GetCheckListsItemAsyncTask.onPostExecute(CheckListActivity.java:261)
适用于//Filter image I with filter kernel:
// 1/9 1/9 1/9
// 1/9 1/9 1/9
// 1/9 1/9 1/9
//I - Input image in pixel ordered BGR format
//image_width - Number of columns of I
//image_height - Number of rows of I
//J - Destination "smoothed" image in BGR format.
//I and J is pixel ordered BGR color format (size in bytes is image_width*image_height*3):
//BRGBRGBRGBRGBR
//BRGBRGBRGBRGBR
//BRGBRGBRGBRGBR
//BRGBRGBRGBRGBR
//
//Limitations:
//I and J must be two separate arrays (in place computation is not supported).
void BgrSmoothing(const unsigned char I[],
int image_width,
int image_height,
unsigned char J[])
{
const int scale = (int)((1.0/9.0)*(1 << 15) + 0.5); //1/9 expanded by 2^15 (add 0.5 for rounding).
const int rounding_ofs = (1 << 14); //0.5 expanded by 2^14
int x, y;
const unsigned char *I0; //Points beginning of row y-1 (in source image I).
const unsigned char *I1; //Points beginning of row y (in source image I).
const unsigned char *I2; //Points beginning of row y+1 (in source image I).
int x0, x1, x2; //x0 - pixel to the left of x1, x1 - center, x2 - pixel to the right of x1
unsigned char *J1; //Points beginning of row y (in destination image J).
//3x3 source blue pixels, 3x3 source green pixels, 3x3 source red pixels.
unsigned char b00, b01, b02, g00, g01, g02, r00, r01, r02;
unsigned char b10, b11, b12, g10, g11, g12, r10, r11, r12;
unsigned char b20, b21, b22, g20, g21, g22, r20, r21, r22;
unsigned char b, g, r; //Destination blue, green and red pixels.
for (y = 0; y < image_height; y++)
{
if (y == 0)
I0 = I; //Handle first row: use row 0 instead of row -1 (row -1 exceeds image bounds).
else
I0 = &I[(y-1)*image_width*3]; //Pointer to beginning of source row above row y in image I.
I1 = &I[y*image_width*3]; //Pointer to beginning of source row y in image I.
if (y == image_height-1)
I2 = &I[y*image_width*3]; //Handle last row: use row image_height-1 instead of row image_height (row image_height exceeds image bounds).
else
I2 = &I[(y+1)*image_width*3]; //Pointer to beginning of source row below row y in image I.
J1 = &J[y*image_width*3]; //Pointer to beginning of destination row in image J.
//Handle first pixel:
for (x = 0; x < image_width; x++)
{
//Multiply x by 3, to convert pixel index to byte index (in BGR forst each pixel is 3 bytes).
x0 = (x == 0) ? (0) : (x-1)*3; //Handle x0 coordinate of first pixel in the row.
x1 = x*3;
x2 = (x == image_width-1) ? (image_width-1)*3 : (x+1)*3; //Handle x2 coordinate of last pixel in the row.
//Load 3x3 blue pixels:
b00 = I0[x0]; b01 = I0[x1], b02 = I0[x2];
b10 = I1[x0]; b11 = I1[x1], b12 = I1[x2];
b20 = I2[x0]; b21 = I2[x1], b22 = I2[x2];
//Load 3x3 green pixels:
g00 = I0[x0+1]; g01 = I0[x1+1], g02 = I0[x2+1];
g10 = I1[x0+1]; g11 = I1[x1+1], g12 = I1[x2+1];
g20 = I2[x0+1]; g21 = I2[x1+1], g22 = I2[x2+1];
//Load 3x3 red pixels:
r00 = I0[x0+2]; r01 = I0[x1+2], r02 = I0[x2+2];
r10 = I1[x0+2]; r11 = I1[x1+2], r12 = I1[x2+2];
r20 = I2[x0+2]; r21 = I2[x1+2], r22 = I2[x2+2];
//Sum all 9 blue elements, all 9 green elements and all 9 red elements (convert to int to avoid overflow).
int sum_b = (int)b00+(int)b01+(int)b02+(int)b10+(int)b11+(int)b12+(int)b20+(int)b21+(int)b22;
int sum_g = (int)g00+(int)g01+(int)g02+(int)g10+(int)g11+(int)g12+(int)g20+(int)g21+(int)g22;
int sum_r = (int)r00+(int)r01+(int)r02+(int)r10+(int)r11+(int)r12+(int)r20+(int)r21+(int)r22;
//b = round(sum_b*(1/9)).
//Because b i positive round(b) = floor(b+0.5).
//Use following computation instead: b = floor((sum_b*(1.0/9.0)*2^15 + 2^14) / 2^15)
b = (unsigned char)((sum_b*scale + rounding_ofs) >> 15); //Destination blue pixel.
g = (unsigned char)((sum_g*scale + rounding_ofs) >> 15); //Destination green pixel.
r = (unsigned char)((sum_r*scale + rounding_ofs) >> 15); //Destination red pixel.
//Store b,g,r elements to destination row J1.
J1[x1] = b;
J1[x1+1] = g;
J1[x1+2] = r;
}
}
}
，但当我连接到我的Epson收据打印机并发送TextBox的内容时，它会将我的标签解释为正常大小的标签 - 而不是6和85的两个标签我在打印机上获得了默认定位标签，这些标签会破坏输出。
任何人都知道我如何在打印机上设置标签位置（或者可能还有其他方式？），以便打印机输出与文本框中的标签相匹配？
答案 0 :(得分：0)
您需要在文本框中设置制表位，并且制表符在您的打印机中停止使用相同的字符数或距离（取决于您是否使用比例或固定字体）。
这里的命令参考应该（大部分）与您的打印机一起使用。收据打印机确实有一些附加功能，如刀杆和可下载的图像，但它应该足够接近。 https://files.support.epson.com/pdf/general/escp2ref.pdf您还可以谷歌查找特定打印机的技术手册，该手册应该包含任何特定代码。
最大的问题是您需要自己生成整个文档并将其作为RAW格式的文档发送到打印机或打印队列，包含所有初始化序列和控制代码，并将其发送到打印机使用RawPrinterHelper（https://support.microsoft.com/en-us/kb/322090）。
它很混乱，但我知道在点阵或热敏收据打印机上获得精确定位，快速输出和良好文本质量的唯一方法。
您可能能够将制表位设置到所需位置，然后在打印机支持时保存配置。问题是，如果使用标准的Epson驱动程序进行打印，打印机将重新初始化并将所有内容呈现为位图，而不是使用打印机字体和您想要的设置进行打印。
你也可以试试＆＃34; Generic＆＃34;打印机驱动程序，其中包含可以配置打印机初始化代码的插槽，但您仍需要包含字体更改/粗体等所需的任何转义序列。在你的文件中。