LR的参数化问题,回放没问题,参数化报错,求救。。。
本帖最后由 Jacks 于 2010-11-10 14:54 编辑LR的参数化问题,回放没问题,参数化报错,求救。。。小弟第一次弄C/S结构
我的脚本:
Action()
{
lrs_create_socket("socket0", "TCP", "LocalHost=0", "RemoteHost=MJW-I-SERVER:9098", LrsLastArg);
lrs_send("socket0", "buf0", LrsLastArg);
lrs_receive("socket0", "buf1", LrsLastArg);
lrs_send("socket0", "buf2", LrsLastArg);
lrs_receive("socket0", "buf3", LrsLastArg);
lrs_create_socket("socket1", "TCP", "LocalHost=0", "RemoteHost=MJW-I-SERVER:9098", LrsLastArg);
lrs_send("socket1", "buf4", LrsLastArg);
lrs_receive("socket1", "buf5", LrsLastArg);
lrs_send("socket1", "buf6", LrsLastArg);
lrs_receive("socket1", "buf7", LrsLastArg);
lrs_close_socket("socket1");
lrs_close_socket("socket0");
lr_think_time(5);
lrs_create_socket("socket2", "TCP", "LocalHost=0", "RemoteHost=MJW-I-SERVER:9098", LrsLastArg);
lrs_send("socket2", "buf8", LrsLastArg);
lrs_receive("socket2", "buf9", LrsLastArg);
lrs_send("socket2", "buf10", LrsLastArg);
lrs_receive("socket2", "buf11", LrsLastArg);
lr_output_message("name:%s",lr_eval_string("<name>"));
lr_output_message("pass:%s",lr_eval_string("<pass>"));
lrs_create_socket("socket3", "TCP", "LocalHost=0", "RemoteHost=MJW-I-SERVER:9098", LrsLastArg);
lrs_send("socket3", "buf12", LrsLastArg);
lrs_receive("socket3", "buf13", LrsLastArg);
lrs_send("socket3", "buf14", LrsLastArg);
lrs_receive("socket3", "buf15", LrsLastArg);
lr_think_time(5);
lrs_close_socket("socket3");
lrs_create_socket("socket4", "TCP", "LocalHost=0", "RemoteHost=MJW-I-SERVER:9098", LrsLastArg);
lrs_send("socket4", "buf16", LrsLastArg);
lrs_receive("socket4", "buf17", LrsLastArg);
lrs_send("socket4", "buf18", LrsLastArg);
lrs_receive("socket4", "buf19", LrsLastArg);
return 0;
}
数据的部分:
send buf10 192
"\x06\xb9\x01"
"l!http://tempuri.org/IUserBLL/Logi ... 8.0.18:9098/UserBLL"
"\x05"
"Login"
"\x13"
"http://tempuri.org/"
"\x02"
"id\bpasswordV"
"\x02"
"\v"
"\x01"
"s"
"\x04"
"\v"
"\x01"
"a"
"\x06"
"V\bD\n"
"\x1e\x00"
"偒"
"\x01"
"D"
"\x1a"
"璠y"
"\x14"
"珬!荂犇裕"
"\x1a"
"傺YD,D*"
"\xab\x14\x01"
"D\f"
"\x1e\x00"
"偒"
"\x03\x01"
"V"
"\x0e"
"B"
"\x05"
"\n"
"\x07"
"B\t"
"\x99\x03"
"<name>" //参数用户名
"\x99\x03"
"<pass>" //参数密码
"\x01\x01\x01"
回放一次没问题,但参数化,迭代2次就报错了:Action.c(30): Error : socket2 - Invalid argument. Error code : 10022.什么原因呢? socket2没有关闭
迭代时又重新创建了,所以报错 回复 2# skyzhu
这个解决了,但我录制一个新加用户,回放也成功了,但数据库里面为什么没有增加呢? 唯一性约束
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