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Use uploaded cfl filename as a default for outputs

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This commit is contained in:
usov_i 2022-04-21 17:43:31 +02:00
parent b26d7005c5
commit 7b70d30578
1 changed files with 7 additions and 3 deletions
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10
pyzebra/app/panel_ccl_prepare.py
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@ -211,8 +211,12 @@ def create():
# run sxtal_refgen for each kvect provided
for i, kvect in enumerate(kvects, start=1):
params["kvect"] = kvect
if open_cfl.filename:
base_fname = f"{os.path.splitext(open_cfl.filename)[0]}_{i}"
else:
base_fname = f"zebra_{i}"
cfl_path = os.path.join(temp_dir, f"zebra_{i}.cfl")
cfl_path = os.path.join(temp_dir, base_fname + ".cfl")
if open_cfl.value:
cfl_template = io.StringIO(base64.b64decode(open_cfl.value).decode())
else:
@ -235,11 +239,11 @@ def create():
print(comp_proc.stdout)
if i == 1: # all hkl files are identical, so keep only one
hkl_fname = f"zebra_{i}.hkl"
hkl_fname = base_fname + ".hkl"
with open(os.path.join(temp_dir, hkl_fname)) as f:
res_files[hkl_fname] = f.read()
mhkl_fname = f"zebra_{i}.mhkl"
mhkl_fname = base_fname + ".mhkl"
with open(os.path.join(temp_dir, mhkl_fname)) as f:
res_files[mhkl_fname] = f.read()