如何在PHP中编写Shell脚本命令并将其链接到下拉菜单

Question

I have a shell scrip command for displaying the output on a web page.

$output =  shell_exec("cd DSSAT46; ./dscsm046 CSCER046 A KSAS8101.WHX  2>&1");

$output=str_replace(" "," ",$output);

$lines = explode("\n", $output);


echo '<pre>';


echo'<table class="fixtable">';

for ($i = 21; $i < sizeof($lines); $i++)
 {
    echo '<tr>';
    $columns = explode("\t", $lines[$i]);
    foreach ($columns as $data) {
        echo '<td>'.$data.'</td>';
    }
    echo '</tr>';
}
echo '</table>';
echo '</pre>';

?> 

I have a drop down menu with values , lets say wheat,sugarcane,paddy.

<form method="POST" action ="name above php file">

<select name="crop">
              <option value="wheat">Wheat</option>
              <option value="sugarcane">Sugarcane</a>
              <option value="paddy">Paddy</a>
        </select> 
</form>

How do I make the shell script command generic so that the "KSAS8101.WHX" changes to something else whenever I select another value from dropdown menu.

For instance if I select option sugarcane then only the value of "KSAS8101.WHX" should change to "UFGA8504.PPX" and the entire command should remain the same.

Answer 1

This is simple, compare POST value:

if ($_POST['crop'] == "wheat"){
    $options =  "KSAS8101.WHX";
} elseif($_POST['crop'] == "sugarcane") {
    $options =  "UFGA8504.PPX";
} elseif($_POST['crop'] == "paddy") {
    $options =  "UFGA8504.PPX";
}

$output =  shell_exec("cd DSSAT46; ./dscsm046 CSCER046 A $options  2>&1");

$output=str_replace(" "," ",$output);

$lines = explode("\n", $output);


echo '<pre>';


echo'<table class="fixtable">';

for ($i = 21; $i < sizeof($lines); $i++)
{
    echo '<tr>';
    $columns = explode("\t", $lines[$i]);
    foreach ($columns as $data) {
        echo '<td>'.$data.'</td>';
    }
    echo '</tr>';
}
echo '</table>';
echo '</pre>';