bash 建立一个字符串
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building a string
提问by Pascal
I have this code:
我有这个代码：
#!/bin/bash
input="./user.cvs"
while IFS=';' read -r f1 f2 f3 f4 f5 f6 f7 f8 f9 f10 f11 f12 f13
do
path="./QRcodes/$f2$f3.png"
vcard="BEGIN:VCARD%0AN;CHARSET=utf-8:$f3;$f2;;$f1;%0AADR;CHARSET=utf-8;INTL;PARCEL;WORK:;;$f10;$f11;;$f12;$f13%0AEMAIL;INTERNET:$f6%0AORG:$f4%0ATEL;WORK:$f8%0ATEL;FAX;WORK:$f9%0ATITLE:$f5%0AURL;WORK:$f7%0AEND:VCARD"
latex=""
encodedVCard=$(echo "$vcard" | sed -e 's/\+/\%2B/g')
url="http://api.qrserver.com/v1/create-qr-code/?size=300x300&data=$encodedVCard"
wget -O "$path" "$url"
latex+="\n \begin{tabular}{ C C } \includegraphics[height=30mm]{graphic.png} & Name \\ \end{tabular}"
echo $latex
done < "$input"
Everything works except the 'echo $latex' always prints the same line instead of multiple times. What am I missing?
一切正常，除了 'echo $latex' 总是打印同一行而不是多次。我错过了什么？
回答by Brian Agnew
If you want to append to a string, just
如果你想附加到一个字符串，只需
latex="$latex newstring"
or
或者
latex=${latex}newstring
You need to be careful that bash doesn't interpret the above as a new var e.g. "$a4" would be interpreted as a variable a4and not as $awith 4appended.
你必须小心的bash不解释上述的新变种如"$a4“将被解释为一个变量a4而不是$a与4追加。
回答by chepner
Several problems:
几个问题：
Backslash expressions are not treated specially inside double quotes, so "\n" is two characters, "\" and "n", not a single newline. Use $'\n' to include an actual newline
echo $latexexpands the variable, but bash performs wordsplitting on any whitespace contained inlatex, including newlines, so the newlines are never seen byecho. You need to quote the variable:echo "$latex".As pointed out by others, you are resetting the variable in each iteration of the loop.
反斜杠表达式不会在双引号内进行特殊处理，因此“\n”是两个字符，“\”和“n”，而不是单个换行符。使用 $'\n' 来包含一个实际的换行符
echo $latex扩展变量，但 bash 对包含在 中的任何空格latex（包括换行符）执行分词，因此换行符永远不会被echo. 您需要引用变量：echo "$latex"。正如其他人所指出的那样，您正在循环的每次迭代中重置变量。
Putting it all together,
把这一切放在一起，
input="./user.cvs"
latex=""
while IFS=';' read -r f1 f2 f3 f4 f5 f6 f7 f8 f9 f10 f11 f12 f13
do
path="./QRcodes/$f2$f3.png"
vcard="BEGIN:VCARD%0AN;CHARSET=utf-8:$f3;$f2;;$f1;%0AADR;CHARSET=utf-8;INTL;PARCEL;WORK:;;$f10;$f11;;$f12;$f13%0AEMAIL;INTERNET:$f6%0AORG:$f4%0ATEL;WORK:$f8%0ATEL;FAX;WORK:$f9%0ATITLE:$f5%0AURL;WORK:$f7%0AEND:VCARD"
#encodedVCard=$(echo "$vcard" | sed -e 's/\+/\%2B/g')
# You can use bash parameter expansion instead of piping into sed
encodedVCard="${vcard//+/%2B}"
url="http://api.qrserver.com/v1/create-qr-code/?size=300x300&data=$encodedVCard"
wget -O "$path" "$url"
latex+=$'\n \begin{tabular}{ C C } \includegraphics[height=30mm]{graphic.png} & Name \\ \end{tabular}'
echo "$latex"
done < "$input"
回答by Barmar
You have
你有
latex=""
in the loop, so it resets it each time. Put that before the beginning of the loop.
在循环中，所以它每次都会重置它。把它放在循环开始之前。
