Question

我从数据库查询中获取信息并将其添加到下拉菜单表单（放在表格中）。查询位于从表单中调用的单独函数中。它将数据库中的信息添加到表中的正确位置，但它不在下拉菜单中。我使用变量$ a，$ b和$ c来测试我的语法，它可以很好地处理这些变量。这是函数调用的问题吗？有什么想法吗？

以下是代码：

<?php
function fill_dropdown(){
include("../secure/database.php");          
$conn = pg_connect(HOST." ".DBNAME." ".USERNAME." ".PASSWORD)
or die('Could not connect: ' . pg_last_error()); //error if could not connect to database

$query = "SELECT country_code, name FROM lab5.country ORDER BY name ASC";

$result = pg_query($query) or die("Unable to execute: " . pg_last_error($conn));

$numRow = 0;
    //results are good so output them to HTML
//echo "test<br />";
while ($line1 = pg_fetch_array($result, null, PGSQL_ASSOC)){
    $counter = 0;
    //echo "test<br />";
    foreach ($line1 as $col_value){ // then add all data for attributes in succeeding columns
        if($counter == 0){
            $code[$numRow] = $col_value;//array($numRow => $col_value);
            echo "\t\t<input type=\"hidden\" name=\"code\" value=\"$code[$numRow]\" />";
            //echo $code[$numRow] . "<br />";
        }
        elseif($counter == 1){
            $country_name[$numRow] = $col_value;
            echo "<option value=$country_name[$numRow]>$country_name[$numRow]</option>";
            //echo $country_name[$numRow] . "<br />";
        }
        $counter++;
    }
    $numRow++;
}
//echo "end test<br />";
}




echo "<table border = \"1\">";
echo "<form method=\"POST\" action=\"exec.php\">";                  //save and cancel buttons
for($i=1; $i<5; $i++)       //building initial table
{
echo "\t<tr>\n";
echo "\t\t<td>";
if($i == 1)
    echo "Name";
elseif($i == 2)
    echo "Country Code";
elseif($i == 3)
    echo "District";
else
    echo "Population";
echo "</td>\n";

echo "<td>\n";

if($i == 1){
    echo "<input type=\"text\" name=\"name\">";
}
elseif($i == 2){
    echo "<select name=\"country_code\">";              //dropdown box
    $c = 0;    //these are just to show that this way works
    $a = "IOT";
    $b = "test2";
    $numRow = 1;
    echo "<option value=\"IOT\">British Indian Ocean Territory</option>";
    echo "<option value=$a>$b</option>";
    //echo "<option value=" . $country_name[$numRow] . ">" . $country_name[$numRow] . "</option>";
    fill_dropdown();
    //echo "<option value=\"Brunei\">Brunei</option>";
    echo "</select>";
}
elseif($i == 3){
    echo "<input type=\"text\" name=\"district\">";
}
else{
    echo "<input type=\"text\" name=\"population\">";
}
}
echo "</td>";
echo "</tr>";
echo "</table>";                                    
echo "\t\t<input type=\"submit\" value=\"Save\" name=\"save\" />";      
echo "<input type=\"button\" value=\"Cancel\" onclick=\"top.location.href='" . $_SERVER['HTTP_REFERER'] . "';\" />\n";
echo "</form>";


?>

Answer 1

看起来你可以替换

while ($line1 = pg_fetch_array($result, null, PGSQL_ASSOC)){
    $counter = 0;
    //echo "test<br />";
    foreach ($line1 as $col_value){ // then add all data for attributes in succeeding columns
        if($counter == 0){
            $code[$numRow] = $col_value;//array($numRow => $col_value);
            echo "\t\t<input type=\"hidden\" name=\"code\" value=\"$code[$numRow]\" />";
            //echo $code[$numRow] . "<br />";
        }
        elseif($counter == 1){
            $country_name[$numRow] = $col_value;
            echo "<option value=$country_name[$numRow]>$country_name[$numRow]</option>";
            //echo $country_name[$numRow] . "<br />";
        }
        $counter++;
    }
    $numRow++;
}

与

while ($row = pg_fetch_assoc($result)) {
    // why do you want this line at all?
    echo "\t\t<input type=\"hidden\" name=\"code\" value=\"$row[country_code]\"/>";
    echo "<option value=\"$row[name]\">$row[name]</option>";
}

我唯一能看到错误的是，没有引用选项的value属性。但是，通过将隐藏输入与选项交错，我不明白你期望实现什么。根据您的$a, $b, $c模板判断，您可能真正想要的是：

while ($row = pg_fetch_assoc($result)) {
    echo "<option value=\"$row[country_code]\">$row[name]</option>";
}

数据库查询信息到下拉菜单

1 个答案: