有人可以幫我弄清楚為什么這個(gè)最基本的代碼在 PHP 中不起作用嗎？多年來(lái)一直使用 PHP，但從未見(jiàn)過(guò)這種情況......因?yàn)槲以?sandbox.onlinephpfunctions.com 網(wǎng)站上對(duì)其進(jìn)行了測(cè)試，所以它在每個(gè)版本的 php 上都是一致的，請(qǐng)參見(jiàn)代碼片段下方的鏈接。我想做的就是將一個(gè)數(shù)字與另一個(gè)數(shù)字進(jìn)行比較，但是當(dāng)數(shù)字包含浮點(diǎn)數(shù)時(shí)，事情會(huì)變得很奇怪。<?php$lside = 490.84;$rside = 237.80 + 222.00 + 31.04;if( $lside == $rside ){   echo "they are equal\n";} else {   echo "not equal\n";   echo "rside: [$rside]\n";   echo "lside: [$lside]\n";}http://sandbox.onlinephpfunctions.com/code/6dee3a97f68a11e67fbaa8e5c157b827ecd47740幫助和感謝！編輯：這是我最終從收到的答案中解決此問(wèn)題并進(jìn)一步研究的方法：<?php$lside = 490.84;$rside = 237.80 + 222.00 + 31.04;if( (string)$lside == (string)$rside ){   echo "they are equal\n";} else {   echo "not equal\n";   echo "rside: [$rside]\n";   echo "lside: [$lside]\n";}因此，要在 PHP 中比較兩個(gè)數(shù)字，一種安全的方法是強(qiáng)制它們成為字符串……看圖。我進(jìn)行了一些額外的測(cè)試，以確保這是真的并且似乎總是有效。所有這些都返回“好”：if( "1" == "1" ) echo "1 - good\n";if( "1.0" == "1" ) echo "2 - good\n";if( "1" == "1.0" ) echo "3 - good\n";if( "1.0" == "1.0" ) echo "4 - good\n";if( "1.0" == "1.00000000" ) echo "5 - good\n";if( "1.000000000000000" == "1.00000000" ) echo "6 - good\n";if( "2" == (string)(.5+.5+1) ) echo "7 - good\n";if( "2.000000000000000" == (string)(1+1) ) echo "8 - good\n";if( "2.000000000000000" == (string)(1.0+1) ) echo "9 - good\n";if( "2.4" == (string)(1.4+1) ) echo "10 - good\n";if( "2.40000" == (string)(1.4+1) ) echo "11 - good\n";if( "2.000000000000000" == (string)(.5+.5+1) ) echo "12 - good\n";if( "1.5" == (string)(.5+.5+.5) ) echo "13 - good\n";if( "1.500000000000000" == (string)(.5+.5+.5) ) echo "14 - good\n";