Question

i 希望在数据库中填入一个表号的清单箱。这部法律为之写,但似乎并未奏效。

<select id="arrays" name="arrays" style="width: 403px;" class="Fieldcell">
<?php
$dbname =  myBase ;

if (!mysql_connect( localhost ,  root ,   )) {
    echo  Could not connect to mysql ;
    exit;
}

$sql = "SHOW TABLES FROM $dbname";
$result = mysql_query($sql);

if (!$result) {
    echo "DB Error, could not list tables
";
    echo  MySQL Error:   . mysql_error();
    exit;
}

$num_tables = mysql_num_rows($result);


for($i=0;$i<$num_tables;$i++)
{

echo "<option value="$row[i]">$row[i]</option>";

}

/*while ($row = mysql_fetch_row($result)) {

echo <option value="$row[0]">$row[0]</option>";

}*/


mysql_free_result($result);
?>
</select>

Answer 1

The commented out section is the section that will make it work. You are using a for loop where you are trying to loop through a variable $row that is never defined. You need to use mysql_fetch_row() to actually grab the data from the result set. Which it looks like you had, but then commented out. Nuke the for lop and uncomment the while loop. Although looks like you have a syntax error in your while loop (missing quotes). Here is what it should look like:

while ($row = mysql_fetch_row($result)) {
    echo "<option value= {$row[0]} >{$row[0]}</option>";
}

页: 1

你们可以保持现在的道路,但除了你需要补充的休息时间外,还要补充。

$row = mysql_fetch_all($result);

Answer 2

<?php
echo  <select id="arrays" style="width: 403px;" class="Fieldcell"> ;
mysql_connect( localhost , root ,  ) or die( Could not connect to mysql );
$result = mysql_query("SHOW TABLES FROM $dbname") or die("DB error, could not list tables<br />MySQL error: ".mysql_error());
while( $r = mysql_fetch_row($result)) {
    echo  <option value=" .$r[0]. "> .$[0]. </option> ;
}
echo  </select> ;
?>

引证。

Answer 3

尝试:

mysql_connect( localhost , root , pass ); //dont forget the correct password
$dbname=???; // insert the database name here
$thequery=mysql_query("SELECT * FROM $dbname");
echo  <select id="arrays" name="arrays" style="width: 403px;" class="Fieldcell"> ;
while($currow=mysql_fetch_array($thequery)) {
  echo  <option value=" .$currow(COLUMN NAME HERE). "> .$currow(COLUMN NAME HERE). </option> ; //insert the column name (the one that has the values inside)
}
echo  </select> ;

一方面,它不是把数据库作为根基的良好想法。使用根基并非安全。

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