我正在从数据库查询中获取信息并将其添加到下拉菜单表单(放置在表中)。 查询位于从窗体中调用的单独函数中。 它将数据库中的信息添加到表上的正确位置,但不在下拉菜单中。 我使用变量 $a、$b 和 $c 来测试我的语法,它对这些变量工作正常。 函数调用有问题吗? 有什么想法吗?
这是代码:
<?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='" . $_SERVER['HTTP_REFERER'] . "';'" />'n";
echo "</form>";
?>
似乎你可以替换
while ($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++;
}
$numRow++;
}
跟
while ($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>";
}
我唯一能看到的错误是,选项的值属性周围没有引号。不过,我不明白您希望通过将隐藏输入与选项交错来实现什么。从您的$a, $b, $c
模板来看,也许您真正想要的是:
while ($row = pg_fetch_assoc($result)) {
echo "<option value='"$row[country_code]'">$row[name]</option>";
}