关于搜索php数组并返回值的问题

Question

Wondering how I can search through the following array for [law] and just return [fine] for that law?

    Array ( [0] => Array ( 
[law] => Unroadworthy Vehicle 
[fine] => 500 
[jail] => 0
[statute] => Any registered vehicle that is lacking headlights, taillights, windshields, has extensive damage, or deemed unsafe to operate, can be considered unroadworthy. Vehicles to do not return a registration shall also be considered unroadworthy. )
            [1] => Array ( 
[law] => Headlights Required 
[fine] => 500 
[jail] => 0 
[statute] => Failure to use headlights after dark or in poorly lit areas or roadways )

..(array continues through 108 unique items)

I have this to loop through and display the law in a dropdown

echo "<select name=\"charge\">";
$strJsonFileContents = file_get_contents("./includes/laws.json");
$issue = json_decode($strJsonFileContents, true);
$arrlength = count($issue);
for ($x = 0; $x < $arrlength; $x++)
    {
    $law = $issue[$x][law];
    echo "<option name=\"law\" value=\"$law\">$law</option>";
    }
echo "</select><br><br>";

I'm actually not going to do the hidden element part.

I want to search through json_decode for [law] and return it's [fine]

Answer 1

Your $law variable has not the expected value and you should get an error like this:

Use of undefined constant law - assumed 'law'

echo "<select name=\"charge\">";
$strJsonFileContents = file_get_contents("./includes/laws.json");
$issue = json_decode($strJsonFileContents, true);
$arrlength = count($issue);
for ($x = 0; $x < $arrlength; $x++)
    {
    $law = $issue[$x]['law'];
    echo "<option name=\"law\" value=\"$law\">$law</option>";
    }
echo "</select><br><br>";

About the hidden [fine] element you can have a look to this question and do something like this:

$('#mySelect').change(function(){
   var id = $(this).find(':selected')[0].id;
   $('#hiddenFineInput').val(id);
})