Question

我需要创建一个类似‘Acme’，‘Umbrella’，‘Waymart’的字符串，以便在java函数中使用。到目前为止，我有：

$info0 = "SELECT DISTINCT Company FROM CETracker";
$rs0=odbc_exec($conn1,$info0);
$count = 1;
while($row = odbc_fetch_array($rs0))    
{
    ${'V'.$count++} = "" . $row['Company'] . "";
}
$categories = "'$V1', '$V2', '$V3'";
echo $categories;

然后我在函数需要公司列表的地方有$categories个回音。这项工作只要有3家公司。但是当添加更多内容时，我将不得不添加‘$V4’，‘$V5’等等。有任何想法吗？感谢。

Answer 1

使用数组而不是变量变量，然后可以implode结果：

$info0 = "SELECT DISTINCT Company FROM CETracker";
$rs0 = odbc_exec($conn1,$info0);
$companies = array();
while($row = odbc_fetch_array($rs0))    {
    $companies[] = "'" . $row['Company'] . "'";
}
$categories = implode(', ', $companies);
echo $categories;

...如果您想继续使用变量变量，请使用循环创建字符串：

$info0 = "SELECT DISTINCT Company FROM CETracker";
$rs0=odbc_exec($conn1,$info0);
$count = 1;
while($row = odbc_fetch_array($rs0))    {
    ${'V'.$count++} = "" . $row['Company'] . "";
}
$categories = '';
for($x = 1; $x <= $count; $x++) {
    if($x > 1) $categories .= ', '; // add the comma
    $categories .= "'" . ${'V'.$x} . "'"; // add the variable variable
}
echo $categories;

Answer 2

请试试这个：

$info0 = "SELECT DISTINCT Company FROM CETracker";
$rs0=odbc_exec($conn1,$info0);
$count = 1;
$str = '';
while($row = odbc_fetch_array($rs0))    
{
    $str .= "'".$row['Company']."',";
}
$str = substr($str,0,strlen($str)-1);
echo $str;

回应变量变量列表

2 个答案: