如何拆分列以修复 php table
How to break a column to fix a php table
我遇到了一个小问题,我无法破解 table,因为我正在寻找这些 table 中的 16 列来创建一个 unicode table...
目的是找到我必须在哪里打破列的线,就像那里一样:
<?php
$columnas = 16;
$filas = 16;
$word= 0;
$contador = 0;
print "<table border=\"1\">\n";
print "<caption>ASCII</caption>\n";
print "<tbody>\n";
print "<tr>\n";
for ($j = 1; $j <= $columnas; $j++){
if ($j%2 == 1){
print "<th>Codigo</th>\n";
}elseif ($j%2 == 0){
print "<th>Valor</th>\n";
}
}
print "</tr>";
for ($i = 1; $i <= $filas; $i++){
for($i = 1; $i <= 50000; $i++,$contador){
$unicodeChar = "&#{$i}";
$contador--;
print "<td>" .$i. "</td>\n";
print "<td>".$unicodeChar."</td> \n";
}
print "</tr>\n";
}
print "</tbody>\n";
print "</table>\n";
?>
enter image description here
你在第一个循环中使用了模数,所以在第二个循环中再次使用它
print "</tr>";
print "<tr>";
for($i = 1; $i <= 50000; $i++){
$unicodeChar = "&#{$i}";
$contador--;
print "<td>$i</td><td>$unicodeChar</td>\n";
if ( $i % $columnas/2 ) == 0 ) {
print "</tr><tr>\n";
}
}
print "</tr>\n";
非常感谢您的帮助,我已经修好了!!
这是我根据您的想法修复的代码:)
´´´´
<?php
$columnas = 16;
$filas = 16;
$word= 0;
$contador = 0;
print "<table border=\"1\">\n";
print "<caption>ASCII</caption>\n";
print "<tbody>\n";
print "<tr>\n";
for ($j = 1; $j <= $columnas; $j++){
if ($j%2 == 1){
print "<th>Codigo</th>\n";
}elseif ($j%2 == 0){
print "<th>Valor</th>\n";
}
}
print "</tr>";
for ($i = 1; $i <= $filas; $i++){
print "<tr>";
for($i = 1; $i <= 50000; $i++,$contador){
$unicodeChar = "&#{$i}";
$contador--;
print "<td>" .$i. "</td>\n";
print "<td>".$unicodeChar."</td> \n";
if (($i % 8 ) == 0 ){
print "</tr><tr>";
}
}
print "</tr>\n";
}
print "</tbody>\n";
print "</table>\n";
?>
´´´´
如Riggsfolly所说,但可能在模数之前将列数除以 2,因为有 2 个 TD/迭代:
if ( $i % ($columnas/2) ) == 0 ) {
print "</tr><tr>\n";
}
我遇到了一个小问题,我无法破解 table,因为我正在寻找这些 table 中的 16 列来创建一个 unicode table... 目的是找到我必须在哪里打破列的线,就像那里一样:
<?php
$columnas = 16;
$filas = 16;
$word= 0;
$contador = 0;
print "<table border=\"1\">\n";
print "<caption>ASCII</caption>\n";
print "<tbody>\n";
print "<tr>\n";
for ($j = 1; $j <= $columnas; $j++){
if ($j%2 == 1){
print "<th>Codigo</th>\n";
}elseif ($j%2 == 0){
print "<th>Valor</th>\n";
}
}
print "</tr>";
for ($i = 1; $i <= $filas; $i++){
for($i = 1; $i <= 50000; $i++,$contador){
$unicodeChar = "&#{$i}";
$contador--;
print "<td>" .$i. "</td>\n";
print "<td>".$unicodeChar."</td> \n";
}
print "</tr>\n";
}
print "</tbody>\n";
print "</table>\n";
?>
enter image description here
你在第一个循环中使用了模数,所以在第二个循环中再次使用它
print "</tr>";
print "<tr>";
for($i = 1; $i <= 50000; $i++){
$unicodeChar = "&#{$i}";
$contador--;
print "<td>$i</td><td>$unicodeChar</td>\n";
if ( $i % $columnas/2 ) == 0 ) {
print "</tr><tr>\n";
}
}
print "</tr>\n";
非常感谢您的帮助,我已经修好了!!
这是我根据您的想法修复的代码:)
´´´´
<?php
$columnas = 16;
$filas = 16;
$word= 0;
$contador = 0;
print "<table border=\"1\">\n";
print "<caption>ASCII</caption>\n";
print "<tbody>\n";
print "<tr>\n";
for ($j = 1; $j <= $columnas; $j++){
if ($j%2 == 1){
print "<th>Codigo</th>\n";
}elseif ($j%2 == 0){
print "<th>Valor</th>\n";
}
}
print "</tr>";
for ($i = 1; $i <= $filas; $i++){
print "<tr>";
for($i = 1; $i <= 50000; $i++,$contador){
$unicodeChar = "&#{$i}";
$contador--;
print "<td>" .$i. "</td>\n";
print "<td>".$unicodeChar."</td> \n";
if (($i % 8 ) == 0 ){
print "</tr><tr>";
}
}
print "</tr>\n";
}
print "</tbody>\n";
print "</table>\n";
?>
´´´´
如Riggsfolly所说,但可能在模数之前将列数除以 2,因为有 2 个 TD/迭代:
if ( $i % ($columnas/2) ) == 0 ) {
print "</tr><tr>\n";
}