LinuxQuestions.org - grabbing data and displaying an image

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grabbing data and displaying an image

i am attempting to grab roughly 40 pieces of data that is a combination of 0 and 1 flags from mysql. based off of the 0 or 1... either pictA or pictB will be displayed.

can someone please assist me in evaluating my code? i believe the logic is incorrect.

PHP Code:




$status = `mysql -u username -ppassword -h x.x.x.x -e 'SELECT help FROM table' database`; 
 
declare -A ARRAY 
 
for ((i=1; i=$status; i++)); 
do 
        echo ${ARRAY[$(i)]} 
done 
 
 
if [${ARRAY[]} = 1 ]; 
then 
       echo "<img src='pictA.gif'>"; 
 
else [${ARRAY[]} = 0 ]; 
       echo "<img src='pictB.gif'>"; 
fi

Ummm ... w/ the square brackets in the if and the condition
tacked onto the else I'm wondering why it would produce anything
at all; that doesn't look like valid PHP.

Cheers,
Tink

apologies, the code is actually written in bash.

Code:

$status = `mysql -u username -ppassword -h x.x.x.x -e 'SELECT help FROM table' database`;



declare -A ARRAY



for ((i=1; i=$status; i++));

do

        echo ${ARRAY[$(i)]}

done





if [${ARRAY[]} = 1 ];

then

      echo "<img src='pictA.gif'>";



else [${ARRAY[]} = 0 ];

      echo "<img src='pictB.gif'>";

fi

What is declare? is that a bash statement

Quote:

Originally Posted by eoc92866 (Post 4149562)

PHP Code:




$status = `mysql -u username -ppassword -h x.x.x.x -e 'SELECT help FROM table' database`; 
 
declare -A ARRAY 
 
for ((i=1; i=$status; i++)); 
do 
        echo ${ARRAY[$(i)]} 
done 
 
 
if [${ARRAY[]} = 1 ]; 
then 
       echo "<img src='pictA.gif'>"; 
 
else [${ARRAY[]} = 0 ]; 
       echo "<img src='pictB.gif'>"; 
fi

first of all $status = will fail as $status will reference a variable, not declare one. It should be status=`statement`

Second you try to assign a index variable using $status. $status is not an integer. it will be a series of strings. First set returned are the column names of the table. next comes the rows of the table.