sh - Argument variables in a shell script -

i tips problem :

i'd run script bash, 1 call suivitest2 launch ./suivitest2 :

#!/bin/sh  echo $1 echo"-----------------------------------" set classpath="" echo"-----------------------------------"  echo $1  jar in lib/*.jar; classpath="$jar;${classpath}"; done  echo using classpath $classpath  /cygdrive/c/program\ files/java/jdk1.7.0_25/bin/java -classpath ${classpath} suivi.test.testsuivi                         conf/suivi.properties $1  echo $1 

but gave me :

$ ./suivitest2 chrp chrp ./suivitest2: line 4: echo----------------------------------- : command not found ./suivitest2: line 6: echo----------------------------------- : command not found classpath= using classpath lib/xmlsec-1.4.3.jar;lib/wss4j-1.5.9.jar;lib/velosurf-2.3-r526.jar;lib/velocity-> > 1.6-dev.jar;lib/suivi.jar;lib/stillness-0.1-dev.jar;lib/mysql-connector-java-5.1.13-bin.jar;lib/mail-1.4.jar;lib/jdom-1.1.jar;lib/dom4j-1.1.jar;lib/commons-net-3.1.jar;lib/commons-logging-1.1.jar;lib/commons-lang-2.2.jar;lib/commons-discovery-0.2.jar;lib/commons-collections-3.2.jar;lib/axis-jaxrpc-1.4.jar;lib/axis-1.4.jar;c:\projects_boxtale\boxtale\suivi\classes\suivi\test ola conf/suivi.properties classpath= classpath= 

have got ideas? tips? if need more informations, please tell me?

you need add 1 or more space after echo split command , arguments:

echo "-----------------------------------"